KC Sinha: Exercise 4.1- Mathem atics Solution Class 12 Chapter 4 प्रतिलोम त्रिकोणमितीय फलन

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TYPE 1

Question 1

Find the value of 
(i) $sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right)$
Sol :
Let $sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right)= \theta$..(i)
⇒$sin \theta =\dfrac{\sqrt{3}}{2}$
∵$\left[sin \dfrac{\pi}{3}=sin \dfrac{\sqrt{3}}{2}\right]$
⇒$sin \theta = sin \dfrac{\pi}{3}$
⇒$\theta =\dfrac{\pi}{3}$..(ii)
On comparing (i) and (ii) equations
⇒$sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right)=\dfrac{\pi}{3}$

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(ii) $tan^{-1} \left(-\dfrac{1}{\sqrt{3}}\right)$
Sol :
⇒Let $tan^{-1} \left(-\dfrac{1}{\sqrt{3}}\right)= \theta$..(i)
⇒$tan \theta =-\dfrac{1}{\sqrt{3}}$
∵$tan \dfrac{\pi}{6}=sin \dfrac{1}{\sqrt{3}}$
⇒$tan \theta =-tan \dfrac{\pi}{6}$
⇒$tan \theta =tan \left(-\dfrac{\pi}{6}\right)$
⇒$\theta =-\dfrac{\pi}{6}$..(ii)
From (i) and (ii) , we get
⇒$tan^{-1}=-\dfrac{\pi}{6}$

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(iii) $cot^{-1} (-\sqrt{3})$
Sol :
Note: cot^-1 (-θ)=π-cot^-1 θ
⇒cot^-1 (-√3)=π-cot^-1 (√3)
⇒$cot^{-1} (-\sqrt{3})=\pi -cot^{-1} \left(cot \dfrac{\pi}{6}\right)$
⇒$cot^{-1} (-\sqrt{3})=\pi-\dfrac{\pi}{6}$
⇒$cot^{-1} (-\sqrt{3})=\dfrac{6\pi-\pi}{6}=\dfrac{5\pi}{6}$

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(vi) $cot^{-1}.cot \dfrac{5\pi}{4}$
Sol :
Let $cot^{-1} . cot\left(\dfrac{5\pi}{4}\right)=\theta$..(i)
⇒$cot \theta=cot \dfrac{5\pi}{4}$
⇒$cot \theta=cot \left(\pi +\dfrac{\pi}{4}\right)$
∵[cot(π+θ)=cotθ]
⇒$cot \theta =cot \dfrac{\pi}{4}$
⇒$\theta=\dfrac{\pi}{4}$..(ii)
From (i) and (ii) , we get
⇒$cot^{-1}. cot \dfrac{5\pi}{4}=\dfrac{\pi}{4}$

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(v) $tan^{-1} \left(tan\dfrac{3\pi}{4}\right)$
Sol :
Let $tan^{-1} \left(tan\dfrac{3\pi}{4}\right)=\theta$..(i)
⇒$tan \theta = tan \dfrac{3\pi}{4}$
⇒$tan \theta = tan \left(\pi -\dfrac{\pi}{4}\right)$
[∵ tan(π-θ)=-tanθ]
⇒$tan \theta=-tan \dfrac{\pi}{4}$
⇒$tan \theta=tan \left(-\dfrac{\pi}{4}\right)$
⇒$\theta=-\dfrac{\pi}{4}$..(ii)
From (i) and (ii) , we get
⇒$tan^{-1}\left(tan \dfrac{3\pi}{4}\right)=-\dfrac{\pi}{4}$

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(vi) $sin^{-1} \dfrac{1}{2}+cos^{-1} \dfrac{1}{2}$
Sol :
Note : $sin^{-1} x+ cos^{-1} x=\dfrac{\pi}{2}$
∴ $sin^{-1} \dfrac{1}{2} + cos^{-1} \dfrac{1}{2} =\dfrac{\pi}{2}$

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(vii) $tan^{-1} \left(tan \dfrac{7\pi}{6}\right)$
Sol :
Let $tan^{-1}\left(tan \dfrac{7\pi}{6}\right)=\theta$..(i)
⇒$tan \theta=tan\dfrac{7\pi}{6}$
⇒$tan \theta=tan\left(\pi+\dfrac{\pi}{6}\right)$
[∵ tan(π+θ)=tanθ]
⇒$tan \theta = tan \dfrac{\pi}{6}$
⇒$\theta =\dfrac{\pi}{6}$..(ii)
From (i) and (ii) , we get
⇒$tan^{-1} \left(tan \dfrac{7\pi}{6}\right)=\dfrac{\pi}{6}$

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(viii) $cos^{-1}. cos \left(\dfrac{13\pi}{6}\right)$
Sol :
Let $cos^{-1}. cos \left(\dfrac{13\pi}{6}\right)=\theta$..(i)
⇒$cos \theta =cos \dfrac{13\pi}{6}$
⇒$cos \theta =cos \left( \pi+\dfrac{7\pi}{6}\right)$
[∵ cos(π+θ)=-cosθ]
⇒$cos \theta=-cos\left(\dfrac{7\pi}{6}\right)$
⇒$cos \theta =-cos \left(\pi+\dfrac{\pi}{6}\right)$
⇒$cos \theta =cos \left(\dfrac{\pi}{6}\right)$
⇒$\theta=\dfrac{\pi}{6}$..(ii)
From (i) and (ii) , we get
⇒$cos^{-1}. cos \left(\dfrac{13\pi}{6}\right)=\dfrac{\pi}{6}$

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(ix) $sin^{-1} \left(sin \dfrac{3\pi}{5}\right)$
Sol :
Let $sin^{-1} \left(sin \dfrac{3\pi}{5}\right)=\theta$..(i)
⇒$sin \theta =sin \dfrac{3\pi}{5}$
⇒$sin \theta =sin \left(\pi -\dfrac{2\pi}{5}\right)$
[∵ sin(π-θ)=sinθ]
⇒$sin \theta=sin \dfrac{2\pi}{5}$
⇒$\theta=\dfrac{2\pi}{5}$..(ii)
From (i) and (ii)
⇒$sin^{-1} \left(sin \dfrac{3\pi}{5}\right)=\dfrac{2\pi}{5}$

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Question 2

(i) tan^-1 (√3)
Sol :
Let tan^-1 (√3)=θ
⇒tanθ=√3
∵$\left[tan \dfrac{\pi}{3}=\sqrt{3}\right]$
⇒$tan \theta=tan \dfrac{\pi}{3}$
⇒$\theta =\dfrac{\pi}{3}$

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(ii) $sin^{-1} \left(-\dfrac{1}{2}\right)$
Sol :
Let $sin^{-1} \left(-\dfrac{1}{2}\right)=\theta$
⇒$sin \theta = -\dfrac{1}{2}$
∵$\left[sin \dfrac{\pi}{6}=\dfrac{1}{2}\right]$
⇒$sin \theta =-sin \dfrac{\pi}{6}$
⇒$sin \theta =sin \left(-\dfrac{\pi}{6}\right)$
⇒$\theta=-\dfrac{\pi}{6}$
⇒$sin^{-1} \left(-\dfrac{1}{2}\right)=-\dfrac{\pi}{6}$

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(iii) tan^-1 (-1)
Sol :
Let tan^-1 (-1)=θ
⇒tanθ=-1
∵$tan\dfrac{\pi}{4}=1$
⇒$\theta=-\dfrac{\pi}{4}$
⇒$tan^{-1} (-1)=-\dfrac{\pi}{4}$

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(iv) cosec^-1 (2)
Sol :
Let cosec^-1 (2)=θ
⇒cosecθ=2
⇒$cosec \theta =cosec \dfrac{\pi}{6}$
⇒$\theta =\dfrac{\pi}{6}$
∴ $cosec^{-1} (2)=\dfrac{\pi}{6}$

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(v) $cos^{-1} \left(-\dfrac{1}{2}\right)$
Sol :
Let $cos^{-1} \left(-\dfrac{1}{2}\right)=\theta$
⇒$cos \theta=-\dfrac{1}{2}$
⇒$cos \theta = cos \dfrac{2\pi}{3}$
∴$cos^{-1}=\dfrac{2\pi}{3}$

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(vi) $cos^{-1} \left(\dfrac{\sqrt{3}}{2}\right)$
Sol :
Let $cos^{-1} \left(\dfrac{\sqrt{3}}{2}\right)=\theta$
⇒$cos \theta=\dfrac{\sqrt{3}}{2}$
⇒$cos \theta=\dfrac{\pi}{6}$
⇒$\theta=\dfrac{\pi}{6}$
∴ $cos^{-1} \left(\dfrac{\sqrt{3}}{2}\right)=\dfrac{\pi}{6}$

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(vii) $cos^{-1} \left(cos \dfrac{2\pi}{3}\right)+sin^{-1} \left(sin \dfrac{2\pi}{3}\right)$
Sol :
Let $cos^{-1} \left(cos \dfrac{2\pi}{3}\right)=\alpha$ , $sin^{-1} \left(sin \dfrac{2\pi}{3}\right)=\beta$
⇒$cos \alpha =cos \dfrac{2\pi}{3}$ , $sin \beta = sin \dfrac{2\pi}{3}$
⇒$\alpha =\dfrac{2\pi}{3}$ ,
$sin \beta =sin \left(\pi -\dfrac{\pi}{3}\right)$
[∵ sin(π-θ)=sinθ]
$sin \beta = sin\dfrac{\pi}{3}$
$\beta = \dfrac{\pi}{3}$
⇒$cos^{-1} \left(cos \dfrac{2\pi}{3}\right)+sin^{-1} \left(sin \dfrac{2\pi}{3}\right)$
⇒α+β
⇒$\dfrac{2\pi}{3}+\dfrac{\pi}{3}=\dfrac{3\pi}{3}=\pi$

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(viii) $cos^{-1} \left(\dfrac{1}{2}\right)-2sin^{-1} \left(-\dfrac{1}{2}\right)$
Sol :
[∵ sin^-1 (-x)=-sin^-1 x]
⇒$cos^{-1} \left(cos \dfrac{\pi}{3}\right)+2sin^{-1} \left(\dfrac{1}{2}\right)$
⇒$cos^{-1} \left(cos \dfrac{\pi}{3}\right)+2sin^{-1} \left(sin\dfrac{\pi}{6}\right)$
⇒$\dfrac{\pi}{3}+\dfrac{2\pi}{6}$
⇒$\dfrac{\pi}{3}+\dfrac{\pi}{3}=\dfrac{2\pi}{3}$

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Question 3

(i) $cos \left[tan^{-1}\left(\dfrac{3}{4}\right)\right]$
Sol :
Let $tan^{-1}\dfrac{3}{4}=\theta$
⇒$tan \theta =\dfrac{3}{4}=\dfrac{p}{b}$
[h=√p²+b²
=√3²+4²
=√25=5]
⇒$cos \theta =\dfrac{b}{h}=\dfrac{4}{5}$
⇒$ \theta =cos^{-1} \dfrac{4}{5}$
⇒$tan^{-1}\dfrac{3}{4}=cos^{-1}\dfrac{4}{5}$
On putting $cos^{-1} \dfrac{4}{5}$ in the place of $tan^{-1}\dfrac{3}{4}$
We get $cos \left[cos^{-1}\dfrac{4}{5}\right]=\dfrac{4}{5}$

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(ii) $cos \left[cos^{-1} \left(\dfrac{\sqrt{3}}{2}\right)+\dfrac{\pi}{6}\right]$
Sol :
⇒$cos \left[\dfrac{\pi}{6}+\dfrac{\pi}{6}\right]$
⇒$cos \left[\dfrac{2\pi}{6}\right]=cos \dfrac{\pi}{3}=\dfrac{1}{2}$

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(iii) $2arc~sin\left(\dfrac{1}{2}\right)+3arc~tan (-1)+2arc~cos\left(-\dfrac{1}{2}\right)$
Sol :
⇒$2sin^{-1} \left(\dfrac{1}{2}\right)+3tan^{-1} (-1)+2cos^{-1} \left(-\dfrac{1}{2}\right)$
⇒$2sin^{-1}\left(sin \dfrac{\pi}{6}\right)-3tan^{-1} (1)+2cos^{-1} \left(cos \dfrac{2\pi}{3}\right)$
⇒$2sin^{-1}\left(sin \dfrac{\pi}{6}\right)-3tan^{-1} \left(tan \dfrac{\pi}{4}\right)+2cos^{-1} \left(cos \dfrac{2\pi}{3}\right)$
⇒$2\times \dfrac{\pi}{6}-\dfrac{3\pi}{4}+2\times \dfrac{2\pi}{3}$
⇒$\dfrac{\pi}{3}-\dfrac{3\pi}{4}+\dfrac{4\pi}{3}=\dfrac{4\pi-9\pi+16\pi}{12}$
⇒$\dfrac{20\pi-9\pi}{12}=\dfrac{11\pi}{12}$

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TYPE 2

Question 4

(i) $tan^{-1} \left(\dfrac{1}{\sqrt{x^2-1}}\right)$ |x|>1
Sol :
Putting x=secθ , then θ=sec^-1 x
Now , $tan^{-1} \left(\dfrac{1}{\sqrt{sec^2 \theta-1}}\right)$
⇒$tan^{-1} \left(\dfrac{1}{\sqrt{tan^{2} \theta}}\right)$
⇒$tan^{-1} \left(\dfrac{1}{tan \theta}\right)$
⇒tan^-1 cotθ
⇒$tan^{-1} . tan \left(\dfrac{\pi}{2}-\theta\right)$
⇒$\dfrac{\pi}{2}-\theta=\dfrac{\pi}{2}-sec^{-1} x$

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(ii) $tan^{-1} \dfrac{\sqrt{1+x^2}-1}{2}$ , x≠0
Sol :
Putting x=tanθ and θ=tan^-1 x
Now $tan^{-1} \dfrac{\sqrt{1+tan^2 \theta}-1}{tan \theta}$
⇒$tan^{-1} \dfrac{\sqrt{sec^2 \theta}-1}{tan \theta}$
⇒$tan^{-1} \dfrac{sec \theta -1}{tan \theta}$
⇒$tan^{-1} \dfrac{\dfrac{1}{cos \theta}-1}{\dfrac{sin \theta}{cos \theta}}$
⇒$tan^{-1} \dfrac{1-cos \theta}{cos \theta} \times \dfrac{cos \theta}{sin \theta}$
⇒$tan^{-1} \dfrac{1- cos \theta}{sin \theta}$
∵$\left[1-cos \theta=2sin^2 \dfrac{\theta}{2}\right]$
∵$sin \theta =2sin \dfrac{\theta}{2}.cos \dfrac{\theta}{2}$
⇒$tan^{-1} \dfrac{2sin^2 \dfrac{\theta}{2}}{2sin \dfrac{\theta}{2}.cos \dfrac{\theta}{2} }$
⇒$tan^{-1} \dfrac{sin \dfrac{\theta}{2}}{cos \dfrac{\theta}{2}}$
⇒$tan^{-1} tan \dfrac{\theta}{2}=\dfrac{\theta}{2}$
⇒$\dfrac{tan^{-1} x}{2}$

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(iii) $tan^{-1} \dfrac{cos x}{1+sinx}$
Sol :
Note: $cos x=\dfrac{1-tan^2 \dfrac{x}{2}}{1+tan^2 \dfrac{x}{2}}$
$sin x=\dfrac{2tan \dfrac{x}{2}}{1+tan^2 \dfrac{x}{2}}$
⇒$tan^{-1}\dfrac{\left(\dfrac{1-tan^2 \frac{x}{2}}{1+tan^2 \frac{x}{2}}\right)}{\left(1+\dfrac{2tan \frac{x}{2}}{1+tan^2 \frac{x}{2}}\right)}$
⇒$tan^{-1} \dfrac{\left(\dfrac{1-tan^2 \frac{x}{2}}{1+tan^2 \frac{x}{2}}\right)}{\left(\dfrac{1+tan^2 \frac{x}{2}+2tan \frac{x}{2}}{1+tan^2 \frac{x}{2}}\right)}$
⇒$tan^{-1} \left[\dfrac{1^2-tan^2 \frac{x}{2}}{\left(1+tan \frac{x}{2}\right)^2}\right]$
⇒$tan^{-1} \dfrac{\left(1-tan \frac{x}{2}\right)\left(1+tan \frac{x}{2}\right)}{\left(1+tan \frac{x}{2}\right)\times\left(1+tan \frac{x}{2}\right)}$
⇒$tan^{-1} \left(\dfrac{1-tan \frac{x}{2}}{1+tan \frac{x}{2}}\right)$
⇒$tan^{-1} \left(\dfrac{tan\frac{\pi}{4}-tan \frac{x}{2}}{1+tan\frac{\pi}{4}.tan \frac{x}{2}}\right)$
$\left[tan (A-B)=\dfrac{tanA-tanB}{1+tanA.tanB}\right]$
⇒$tan^{-1} . tan\left(\dfrac{x}{4}-\dfrac{x}{2}\right)$
⇒$\dfrac{x}{4}-\dfrac{x}{2}$

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(iv) $cot^{-1} \left[\dfrac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\right]$
Sol :
⇒$cot^{-1}\left[\dfrac{1+\dfrac{\sqrt{1-sinx}}{\sqrt{1+sinx}}}{1-\dfrac{\sqrt{1-sinx}}{\sqrt{1+sinx}}}\right]$
Note : $1-cos \theta =2sin^2 \dfrac{\theta}{2}$ $1+cos \theta =2cos^2 \dfrac{\theta}{2}$
Now $\sqrt{\dfrac{1-sinx}{1+sinx}}=\sqrt{\dfrac{1-cos\left(\frac{x}{2}-x\right)}{1+cos\left(\frac{x}{2}-x\right)}}$ $\sqrt{\dfrac{2 sin^2 \left(\frac{\pi}{4}-\frac{x}{2}\right)}{2cos^2 \left(\frac{\pi}{4}-\frac{x}{2}\right)}}$
⇒$\dfrac{sin\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)}{cos\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)}=tan \left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$
∴$\sqrt{\dfrac{1-sin x}{1+sin x }}=tan \left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$
⇒$cot^{-1}\left(\dfrac{1+tan\left(\frac{\pi}{4}-\frac{x}{2}\right)}{1-tan \left(\frac{\pi}{4}-\frac{x}{2}\right)}\right)$
∵$\dfrac{tan A+tanB}{1-tanA.tanB}=tan(A+B)$
⇒$cot^{-1}\left(\dfrac{tan\frac{\pi}{4}+tan\left(\frac{\pi}{4}-\frac{x}{2}\right)}{1-tan \frac{\pi}{4}.tan\left(\frac{\pi}{4}-\frac{x}{2}\right)}\right)$
⇒$cot^{-1} . tan\left(\dfrac{\pi}{4}+\dfrac{\pi}{4}-\dfrac{x}{2}\right)$
⇒$cot^{-1} cot \left(\dfrac{\pi}{2}-\left(\dfrac{\pi}{4}+\dfrac{\pi}{4}-\dfrac{x}{2}\right)\right)$
⇒$\dfrac{\pi}{2}-\dfrac{\pi}{4}-\dfrac{\pi}{4}+\dfrac{x}{2}=\dfrac{x}{2}$

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(v) $cos^{-1} \sqrt{\dfrac{\sqrt{1+x^2}+1}{2\sqrt{1+x^2}}}$
Sol :
Putting x=tanθ , then θ=tan^-1 x
Now $cos^{-1} \sqrt{\dfrac{\sqrt{1+tan^2 \theta}+1}{2\sqrt{1+tan^2 \theta}}}$
[∵ 1+tan²θ=sec²θ]
⇒$cos^{-1} \sqrt{\dfrac{sec\theta+1}{2 sec \theta}}$
⇒$cos^{-1} \sqrt{\dfrac{1+cos \theta}{\dfrac{2}{cos \theta}\times cos \theta}}$
⇒$cos^{-1} \sqrt{\dfrac{1+cos \theta}{2}}$
⇒$cos^{-1} \sqrt{\dfrac{2cos^2 \frac{\theta}{2}}{2}}$
⇒$cos^{-1} . cos\dfrac{\theta}{2}$
⇒$\dfrac{\theta}{2}=\dfrac{tan^{-1}x}{2}$

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(vi) $tan^{-1} \left(\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$
Sol :
Putting x=cosθ , then $\theta=\dfrac{1}{2}cos^{-1}x$
Now , $tan^{-1}\left(\dfrac{\sqrt{1+cos2\theta}-\sqrt{1-cos2\theta}}{\sqrt{1+cos2\theta}+\sqrt{1-cos2\theta}}\right)$
⇒$tan^{-1}\left(\dfrac{\sqrt{2cos^2\theta}-\sqrt{2sin^2\theta}}{\sqrt{2cos^2\theta}+\sqrt{2sin^2\theta}}\right)$
⇒$tan^{-1} \dfrac{\sqrt{2}cos \theta-\sqrt{2}sin \theta}{\sqrt{2}cos \theta+\sqrt{2}sin \theta}$
⇒$tan^{-1} \dfrac{\sqrt{2}(cos \theta-sin \theta)}{\sqrt{2}(cos \theta+sin \theta)}$
[dividing by cosθ]
⇒$tan^{-1}\left(\dfrac{1-tan\theta}{1+tan \theta}\right)$
⇒$tan^{-1}\left(\dfrac{tan \dfrac{\pi}{4}-tan\theta}{1+tan \dfrac{\pi}{4} .tan \theta}\right)$
⇒$tan^{-1} tan \left(\dfrac{\pi}{4}-\theta\right)$
⇒$\dfrac{\pi}{4}-\theta$
⇒$\dfrac{\pi}{4}-\dfrac{1}{2} cos^{-1}x$

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TYPE 3

Question 5

Prove that:
(i) $tan^{-1} \dfrac{2}{11}+tan^{-1} \dfrac{7}{24}$
Sol :
Note: $tan^{-1} A+tan^{-1} B=tan^{-1}\left(\dfrac{A+B}{1-AB}\right)$
Now , L.H.S = $tan^{-1} \dfrac{2}{11}+tan^{-1} \dfrac{7}{24}=tan^{-1} \left(\dfrac{\dfrac{2}{11}+\dfrac{7}{24}}{1-\dfrac{2}{11}\times \dfrac{7}{24}}\right)$
⇒$tan^{-1} \left(\dfrac{\dfrac{48+77}{264}}{1-\dfrac{14}{264}}\right)$
⇒$tan^{-1} \left(\dfrac{125}{250}\right)=tan^{-1} \left(\dfrac{1}{2}\right)$ =R.H.S
Hence Proved

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(ii) $tan^{-1} \dfrac{1}{2}+tan^{-1} \dfrac{1}{3}=tan^{-1} \dfrac{3}{5}+tan^{-1} \dfrac{1}{4}=\dfrac{\pi}{4}$
Sol :
Note : $tan^{-1} A + tan^{-1} B = tan^{-1} \left(\dfrac{A+B}{1-AB}\right)$
Now L.H.S=$tan^{-1}\dfrac{1}{2}+tan^{-1}\dfrac{1}{3}=tan^{-1}\left(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}}{1-\dfrac{1}{2}\times \dfrac{1}{3}}\right)$
⇒$tan^{-1} \left(\dfrac{\dfrac{3+2}{6}}{\dfrac{6-1}{6}}\right)$
⇒$tan^{-1} \left(\dfrac{5}{5}\right)=tan^{-1} (1)$
⇒$tan^{-1} \left(tan \dfrac{\pi}{4}\right)=\dfrac{\pi}{4}$ R.H.S..(i)
Again , L.H.S=$tan^{-1} \dfrac{3}{5}+tan^{-1} \dfrac{1}{4}=tan^{-1} \left(\dfrac{\dfrac{3}{5}+\dfrac{1}{4}}{1-\dfrac{3}{5}\times \dfrac{1}{4}}\right)$
⇒$tan^{-1} \left(\dfrac{\dfrac{12+5}{20}}{1-\dfrac{3}{20}}\right)$
⇒$tan^{-1} \left(\dfrac{\dfrac{17}{20}}{\dfrac{20-3}{20}}\right)$
⇒$tan^{-1} \left(\dfrac{17}{17}\right)=tan^{-1} (1)$
⇒$tan^{-1} \left(tan \dfrac{\pi}{4}\right)=\dfrac{\pi}{4}$..(ii)
From (i) and (ii) , we get
$tan^{-1} \dfrac{1}{2}+tan^{-1} \dfrac{1}{3}=tan^{-1} \dfrac{3}{5}+tan^{-1} \dfrac{1}{4}=\dfrac{\pi}{4}$
Hence Proved

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(iii) $tan^{-1} \dfrac{2a-b}{b\sqrt{3}}+tan^{-1} \dfrac{2b-a}{a\sqrt{3}}=\dfrac{\pi}{3}$
Sol :
L.H.S =$tan^{-1} \dfrac{2a-b}{b\sqrt{3}}+tan^{-1} \dfrac{2b-a}{a\sqrt{3}}=\dfrac{\pi}{3}$
⇒$tan^{-1} \left(\dfrac{\dfrac{2a-b}{b\sqrt{3}}+\dfrac{2b-a}{a\sqrt{3}}}{1-\dfrac{2a-b}{b\sqrt{3}}\times \dfrac{2b-a}{a\sqrt{3}}}\right)$
⇒$tan^{-1} \left(\dfrac{\dfrac{a\sqrt{3}(2a-b)+b\sqrt{3}(2b-a)}{b\sqrt{3}.a\sqrt{3}}}{1-\dfrac{(2a-b)(2b-a)}{b\sqrt{3}.a\sqrt{3}}}\right)$
⇒$tan^{-1} \left(\dfrac{\dfrac{2\sqrt{3}a^2-ab\sqrt{3}+2\sqrt{3}b^2-ab\sqrt{3}}{b\sqrt{3}.a\sqrt{3}}}{\dfrac{b\sqrt{3}.a\sqrt{3}-(4ab-2a^2-2b^2+ab)}{b\sqrt{3}.a\sqrt{3}}}\right)$
⇒$tan^{-1} \left(\dfrac{2\sqrt{3}a^2+2\sqrt{3}b^2-2ab\sqrt{3}}{3ab-4ab+2a^2+2b^2-ab}\right)$
⇒$tan^{-1} \left(\dfrac{2\sqrt{3}(a^2+b^2-ab)}{2(a^2+b^2-ab)}\right)$
⇒$tan^{-1} (\sqrt{3})=tan^{-1} (tan \dfrac{\pi}{3})=\dfrac{\pi}{3}$ =R.H.S
Hence Proved

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(iv) $tan^{-1} \dfrac{2}{5}+tan^{-1} \dfrac{1}{3}+tan^{-1} \dfrac{1}{12}=\dfrac{\pi}{4}$
Sol :
L.H.S=$tan^{-1} \dfrac{2}{5}+tan^{-1} \dfrac{1}{3}+tan^{-1} \dfrac{1}{12}$
⇒$tan^{-1} \left(\dfrac{\frac{2}{5}+\frac{1}{3}}{1-\frac{2}{5} \times \frac{1}{3}}\right)+tan^{-1} \dfrac{1}{12}$
⇒$tan^{-1} \left(\dfrac{\dfrac{6+5}{15}}{\dfrac{15-2}{15}}\right)+tan^{-1} \dfrac{1}{12}$
⇒$tan^{-1} \left(\dfrac{11}{13}\right)+tan^{-1} \dfrac{1}{12}$
⇒$tan^{-1} \left(\dfrac{\frac{11}{13}+\frac{1}{12}}{1-\frac{11}{13}\times \frac{1}{12}}\right)$
⇒$tan^{-1} \left(\dfrac{\frac{132+13}{156}}{\frac{156-11}{156}}\right)$
⇒$tan^{-1} \left(\dfrac{145}{145}\right)=tan^{-1}$
⇒$tan^{-1} \left(tan\dfrac{\pi}{4}\right)=\dfrac{\pi}{4}$=R.H.S proved

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(v)  $2tan^{-1} \dfrac{1}{5}+tan^{-1} \dfrac{1}{4}=tan^{1} \dfrac{32}{45}$
Sol :
∵$\left(2tan^{-1} x=tan^{-1}\dfrac{2x}{1-x^2}\right)$
⇒$2tan^{-1} \dfrac{1}{5}=tan^{-1} \left(\dfrac{2\dfrac{1}{5}}{1-\dfrac{1}{5^2}}\right)$
⇒$tan^{-1} \dfrac{\dfrac{2}{5}}{\dfrac{25-1}{25}}=tan^{-1} \dfrac{10}{24}$
⇒$tan^{-1} \dfrac{5}{12}$
∴$2tan^{-1} \dfrac{1}{5}=tan^{-1} \dfrac{5}{12}$
Now, L.H.S $2tan^{-1} \dfrac{1}{5}+tan^{-1} \dfrac{1}{4}$
⇒$tan^{-1} \dfrac{5}{12}+tan^{-1} \dfrac{1}{4}$
⇒$tan^{-1} \left(\dfrac{\frac{5}{12}+\frac{1}{4}}{1-\frac{5}{12}\times \frac{1}{4}}\right)$
⇒$tan^{-1} \left(\dfrac{\frac{20+12}{48}}{\frac{48-5}{48}}\right)$
⇒$tan^{-1} \left(\dfrac{32}{43}\right)$ R.H.S proved

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(vi) $2tan^{-1} \dfrac{1}{2}+tan^{-1} \dfrac{1}{7}=tan^{-1} \dfrac{31}{17}$
Sol :
Note :$\left(2tan^{-1} x=tan^{-1}\dfrac{2x}{1-x^2}\right)$
⇒$2 tan^{-1} \dfrac{1}{2}=tan^{-1} \dfrac{2\times \frac{1}{2}}{1-\frac{1}{2^2}}$
$=tan^{-1} \dfrac{1}{1-\frac{1}{4}}=tan^{-1} \dfrac{4}{4-1}$
$=tan^{-1} \dfrac{4}{3}$
Now , L.H.S =$2tan^{-1} \dfrac{1}{2}+tan^{-1} \dfrac{1}{7}$
⇒$tan^{-1} \dfrac{4}{3}+tan^{-1} \dfrac{1}{7}$
⇒$tan^{-1} \left(\dfrac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}\times \frac{1}{7}}\right)$
⇒$tan^{-1} \left(\dfrac{\frac{28+3}{21}}{\frac{21-4}{21}}\right)$
⇒$tan^{-1} \dfrac{31}{17}$ R.H.S proved

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(vii) tan^-1 1+tan^-1 2+tan^-1 3=π$=2(tan^{-1} 1+tan^{-1}\dfrac{1}{2}+tan^{-1}\dfrac{1}{3})$
Sol :
L.H.S =tan^-1 1+tan^-1 2+tan^-1 3
=$tan^{-1} \left(\dfrac{1+2}{1-1\times2}\right)+tan^{-1}3$
=$tan^{-1} \left(\dfrac{3}{1-2}\right)+tan^{-1}3$
=$tan^{-1} \left(\dfrac{3}{-1}\right)+tan^{-1}3$
=tan^-1 3 - tan^-1 3
=$tan^{-1} \left(\dfrac{3-3}{1+3\times 3}\right)$
=$tan^{-1} \left(\dfrac{0}{10}\right)=tan^{-1}0$
=tan^-1(tanπ)=π=R.H.S..(i)
Again , R.H.S=$2\left(tan^{-1} 1+tan^{-1} \dfrac{1}{2}+tan^{-1} \dfrac{1}{3}\right)$
=$2\left\{tan^{-1}\left(\dfrac{1+\frac{1}{2}}{1-1\times \frac{1}{2}}\right)+tan^{-1} \dfrac{1}{3}\right\}$
=$2\left\{tan^{-1}\left(\dfrac{\frac{3}{2}}{\times \frac{2-1}{2}}\right)+tan^{-1} \dfrac{1}{3}\right\}$
=$2 \left\{tan^{-1} \dfrac{3}{1}+tan^{-1} \dfrac{1}{3}\right\}$
=$2 \left\{tan^{-1} \left(\dfrac{3+\frac{1}{3}}{1-3\times \frac{1}{3}}\right) \right\}$
=$2 \left\{tan^{-1} \left(\dfrac{\frac{10}{3}}{\frac{1-1}{3}}\right) \right\}$
=$2 \left\{tan^{-1} \left(\dfrac{10}{0}\right) \right\}$
=$2 \left\{ tan^{-1} (undefined)\right\}=2\left\{tan^{-1} tan \dfrac{\pi}{2}\right\}$
=$2 \dfrac{\pi}{2}=\pi$ L.H.S ..(ii)
From (i) and (ii)
⇒tan^-1 1+tan^-1 2+tan^-1 3=π$=2(tan^{-1} 1+tan^{-1}\dfrac{1}{2}+tan^{-1}\dfrac{1}{3})$

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(viii) $tan^{-1} \left[2cos \left(2sin^{-1} \dfrac{1}{2}\right)\right]$
Sol :
Given , $tan^{-1} \left[2cos \left(2sin^{-1} \dfrac{1}{2}\right)\right]$
=$tan^{-1} \left[2cos (2sin^{-1} sin \dfrac{\pi}{6})\right]$
=$tan^{-1} \left[2cos \left(2 \dfrac{\pi}{6}\right)\right]$
=$tan^{-1} \left[2cos  \dfrac{\pi}{3}\right] =\tan^{-1} \left[2\times \dfrac{1}{2}\right]$
=$tan^{-1} (1)=tan^{-1} \left(tan \dfrac{\pi}{4}\right)=\dfrac{\pi}{4}$ R.H.S
proved

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(ix) $tan^{-1} \left(sin^{-} \dfrac{3}{5}+cot^{-1} \dfrac{3}{2}\right)=\dfrac{17}{6}$
Sol :
Let $sin^{-1} \dfrac{3}{5}=\alpha$ and $cot^{-1} \dfrac{3}{2}=\beta$
⇒$sin \alpha =\dfrac{3}{5} , cot \beta =\dfrac{3}{2}$
⇒$tan \alpha =\dfrac{3}{4} , tan \beta =\dfrac{2}{3}$
⇒$\alpha=tan^{-1} \dfrac{3}{4} , \beta=tan^{-1} \dfrac{2}{3}$
⇒$sin^{-1} \dfrac{3}{5}=tan^{-1} \dfrac{3}{4} , cot^{-1} \dfrac{3}{2}=tan^{-1} \dfrac{2}{3}$
Now , L.H.S=$tan \left(sin^{-1} \dfrac{3}{5}+cot^{-1} \dfrac{3}{2}\right)$
$=tan\left(tan^{-1} \dfrac{3}{4}+tan^{-1} \dfrac{2}{3}\right)$
$=tan \left(tan^{-1} \left(\dfrac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\times \frac{2}{3}}\right)\right)$
$=tan \left(tan^{-1} \left(\dfrac{\frac{9+8}{12}}{\frac{12-6}{12}}\right)\right)$
=$tan \left(tan^{-1} \dfrac{17}{6}\right)=\dfrac{17}{6}$ R.H.S
proved

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(x) $tan^{-1} \left(\dfrac{1}{2} sin^{-1} \dfrac{3}{4}\right)=\dfrac{4-\sqrt{7}}{3}$
Sol :
Let $\left(\dfrac{1}{2} sin^{-1} \dfrac{3}{4}\right)=\theta$
⇒$sin^{-1} \dfrac{3}{4}=2\theta$
⇒$sin^{-1} 2\theta=\dfrac{3}{4}$
⇒$\dfrac{2tan \theta}{1+tan^{2}\theta}=\dfrac{3}{4}$
⇒8 tanθ=3(1+tan² θ)
⇒8 tanθ=3+3tan² θ)
⇒3tan² θ-8tanθ+3=0
Let tanθ=x , then 3x²-8x+3=0
Discriminant (D)=b²-4ac
=(-8)²-4×3×3
=64-36=28>0

$x=\dfrac{-b \pm \sqrt{D}}{2a}=\dfrac{-(-8) \pm \sqrt{28}}{2\times3}$
$x=\dfrac{8 \pm \sqrt{4\times 7}}{2\times 3}=\dfrac{8 \pm 2\sqrt{7}}{2\times 3}$
$x=\dfrac{2(4 \pm \sqrt{7})}{2\times 3}=\dfrac{4 \pm \sqrt{7}}{3}$
$x=\dfrac{4+\sqrt{7}}{3}$ or $x=\dfrac{4-\sqrt{7}}{3}$
⇒$tan\theta=\dfrac{4-\sqrt{7}}{3}$
⇒$\theta=tan^{-1}\left(\dfrac{4-\sqrt{7}}{3}\right)$
⇒$\dfrac{1}{2} sin^{-1} \dfrac{3}{4}=\theta=tan^{-1} \left(\dfrac{4-\sqrt{7}}{3}\right)$
⇒$tan\left(tan^{-1} \left(\dfrac{4-\sqrt{7}}{3}\right)\right)=\dfrac{4-\sqrt{7}}{3}$ R.H.S
Proved

Question 6

Prove that
(i) $tan^{-1} x+cot^{-1} y=tan^{-1} \dfrac{xy+1}{y-x}$
Sol :
L.H.S=tan^-1 x + cot^-1 y
∵$\left[cot^{-1} x =tan^{-1} \dfrac{1}{x}\right]$
=$tan^{-1} x+tan^{-1}\dfrac{1}{y}$
=$tan^{-1} \left(\dfrac{x+\dfrac{1}{y}}{1-x\times \dfrac{1}{y}}\right)$
=$tan^{-1} \left( \dfrac{\dfrac{xy+1}{y}}{\dfrac{y-x}{y}} \right)$
=$tan^{-1} \dfrac{xy+1}{y-x}$ R.H.S

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(ii) tan^-1 x + cot^-1(1+x) = tan^-1(1+x+x²)
Sol :
L.H.S= tan^-1 x+cot^-1 (1+x)
=$tan^{-1} x+tan^{-1} \dfrac{1}{1+x} $
=$tan^{-1} \left(\dfrac{x+\dfrac{1}{x}}{1-x\times \dfrac{1}{1+x}}\right)$
=$tan^{-1} \left(\dfrac{\dfrac{x(1+x)+1}{1+x}}{\dfrac{1+x-x}{1+x}}\right)$
=$tan^{-1} \dfrac{x+x^2+1}{1}$
= tan^-1(1+x+x²) R.H.S
Proved

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(iii) $tan^{-1} \dfrac{1}{x+y} + tan^{-1} \dfrac{1}{x^2+xy+1}=cot^-{1} x$
Sol :
L.H.S=$tan^{-1} \dfrac{1}{x+y}+tan^{-1} \dfrac{y}{x+xy+1}$
=$tan^{-1}  \left(\dfrac{\dfrac{1}{x+y}+\dfrac{y}{x^2+xy+1}}{1-\dfrac{1}{x+y}\times \dfrac{y}{x^2+xy+1}}\right)$
=$tan^{-1} \dfrac{\left(\dfrac{x^2+xy+1+xy+y^2}{(x+y)(x^2+xy+1)}\right)}{\left(\dfrac{(x+y)(x^2+xy+1)-y}{(x+y)(x^2+xy+1)}\right)}$
=$tan^{-1} \dfrac{x^2+xy+1+xy+y^2}{x^3+xy^2+x+yx^2+xy^2+y-y}$
=$tan^{-1} \dfrac{(x^2+y^2+2xy+1)}{(x^3+xy^2+2x^2y+x)}$
=$tan^{-1} \dfrac{(x^2+y^2+2xy+1)}{(x(x^2+y^2+2xy+1))}$
=$tan^{-1} \dfrac{1}{x}=cot^{-1} x$R.H.S
proved

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(iv) 2 cot^-1 5+cot^-1 7+2 cot^-18=π/4
Sol :
L.H.S=2 cot^-1 5 + cot^-17+2 cot^-18
=2 cot^-1 5 +2 cot^-18+ cot^-17
=2(cot^-1 5+cot^-1 8)+cot^-1 7
=$\left(tan^{-1} \dfrac{1}{5}+tan^{-1} \dfrac{1}{8}\right)+tan^{-1} \dfrac{1}{7}$
=$2\left(tan^{-1} \dfrac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5}\times \frac{1}{8}}\right)+tan^{-1} \dfrac{1}{7}$
=$2 tan^{-1}\left(\dfrac{\dfrac{13}{40}}{\dfrac{39}{40}}\right)+tan^{-1} \dfrac{1}{7}$
=$2 tan^{-1} \dfrac{1}{3}+tan^{-1} \dfrac{1}{7}$
=$tan^{-1} \dfrac{2\times \dfrac{1}{3}}{1-\dfrac{1}{9}}+tan^{-1} \dfrac{1}{7}$
=$tan^{-1} \left(\dfrac{\frac{2}{9}}{\frac{8}{9}}\right)+tan^{-1} \dfrac{1}{7}$
=$tan^{-1} \left(\dfrac{3}{4}\right)+tan^{-1} \dfrac{1}{7}$
=$tan^{-1} \left(\dfrac{\dfrac{3}{4}+\dfrac{1}{7}}{1-\dfrac{3}{4}\times\dfrac{1}{7}}\right)$
=$tan^{-1} \dfrac{\dfrac{25}{28}}{\dfrac{25}{28}}=tan^{-1} 1$
=$tan^{-1} tan \dfrac{\pi}{4}= \dfrac{\pi}{4}$ R.H.S
Proved

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(v) $tan^{-1} \left(\dfrac{x}{8}\right)-tan^{-1}\left(\dfrac{x-y}{x+y}\right)=\dfrac{\pi}{4}$
Sol :
L.H.S=$tan^{-1} \left(\dfrac{x}{y}\right)-tan^{-1} \left(\dfrac{x-y}{x+y}\right)$
=$tan^{-1} \left\{\dfrac{\frac{x}{y}-\frac{x-y}{x+y}}{1+\frac{x}{y} \times \frac{x-y}{x+y}}\right\}$
=$tan^{-1} \dfrac{\left(\dfrac{x(x+y)-y(x-y)}{y(x+y)}\right)}{\left(\dfrac{y(x+y)+x(x-y)}{y(x+y)}\right)}$
=$tan^{-1} \dfrac{[x^2+xy-yx+y^2]}{[yx+y^2+x^2-yx]}$
=$tan^{-1} \dfrac{[x^2+y^2]}{[x^2+y^2]}=tan^{-1} 1$
=$tan^{-1} tan \dfrac{\pi}{4}=\dfrac{\pi}{4}$ R.H.S

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Question 7

Prove that :
(i) $tan^{-1} \dfrac{a-b}{1+ab}+tan^{-1} \dfrac{b-c}{1+bc}+tan^{-1} \dfrac{c-a}{1+ca}=0$ ab>-1 , bc>-1 , ca>-1
Sol :
L.H.S=$tan^{-1} \dfrac{a-b}{1+ab}+tan^{-1} \dfrac{b-c}{1+bc}+tan^{-1} \dfrac{c-a}{1+ca}$
=tan^-1 a-tan^-1 b+tan^-1 b-tan^-1 c+tan^-1 c-tan^-1 a
=tan^-1 a-tan^-1 a-tan^-1 b+tan^-1 b-tan^-1 c+tan^-1 c
=0 =R.H.S 
proved

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(ii) $tan^{-1} \dfrac{a^3-b^3}{1+a^3b^3}+tan^{-1} \dfrac{c^3-a^3}{1+b^3c^3}+tan^{-1} \dfrac{c^3-a^3}{1+c^3a^3}=0$
Sol :
L.H.S=$tan^{-1} \dfrac{a^3-b^3}{1+a^3b^3}+tan^{-1} \dfrac{c^3-a^3}{1+b^3c^3}+tan^{-1} \dfrac{c^3-a^3}{1+c^3a^3}$
=tan^-1 a³-tan^-1 b³+tan^-1 b³-tan^-1 c³+tan^-1 c³-tan^-1 a³
=tan^-1 a³-tan^-1 a³-tan^-1 b³+tan^-1 b³-tan^-1 c³+tan^-1 c³
=0 R.H.S
proved

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Question 8

(i) $sin^{-1} \dfrac{3}{5}+sin^{-1} \dfrac{8}{17}=sin^{-1} \dfrac{77}{85}$
Sol :
L.H.S=$=\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{8}{17}$

$=\sin ^{-1}\left[\frac{3}{5} \cdot \sqrt{1-\frac{8^{2}}{17^{2}}}+\frac{8}{17} \cdot \sqrt{1-\left(\frac{3}{5}\right)^{2}}\right]$

=$\sin ^{-1}\left[\frac{3}{5} \times \frac{\sqrt{17^{2}-8^{2}}}{17}+\frac{8}{17} \cdot \frac{\sqrt{5^{2}-3^{2}}}{5}\right]$

=$\sin ^{-1}\left[\frac{3}{5} \times \frac{\sqrt{289-64}}{17}+\frac{8}{17} \cdot \frac{\sqrt{25-9}}{5}\right]$

$=\sin ^{-1}\left[\frac{3}{5} \times \frac{\sqrt{225}}{17}+\frac{8}{17} \cdot \frac{\sqrt{16}}{5}\right]$

$=\sin ^{-1}\left[\frac{3}{5} \times \frac{15}{17}+\frac{8}{17} \times \frac{4}{5}\right]$

$=\sin ^{-1}\left[\frac{45}{85}+\frac{32}{85}\right]$

$=\sin ^{-1}\left[\frac{77}{85}\right]$=R.H.S proved

(ii) $\cos ^{-1} \frac{3}{5}+\cos ^{-1} \frac{12}{13}+\cos ^{-1} \frac{63}{65}=\frac{\pi}{2}$

Sol :

L.H.S$=\cos ^{-1} \frac{3}{5}+\cos ^{-1} \frac{12}{13}+\cos ^{-1} \frac{63}{65}$

$=\cos ^{-1}\left(\frac{3}{5} \times \frac{12}{13}-\sqrt{1-\left(\frac{3}{5}\right)^{2}} \cdot \sqrt{1-\left(\frac{12}{13}\right)^{2}}\right)+\cos ^{-1} \frac{63}{65} $

$=\cos ^{-1}\left(\frac{36}{65}-\sqrt{\frac{5^{2}-3^{2}}{5^{2}}} \cdot \sqrt{\frac{13^{2}-12^{2}}{13^{2}}}\right)+\cos ^{-1} \frac{63}{65} $

$=\cos ^{-1}\left(\frac{36}{65}-\frac{\sqrt{25-9}}{5} \cdot \frac{\sqrt{169-144}}{13}\right)+\cos ^{-1} \frac{63}{65}$

$=\cos ^{-1}\left(\frac{36}{65}-\frac{4}{5} \times \frac{5}{13}\right)+\cos ^{-1} \frac{63}{65}$

$=\cos ^{-1}\left(\frac{36}{65}-\frac{20}{65}\right)+\cos ^{-1} \frac{63}{65}$

$=\cos ^{-1} \frac{16}{65}+\cos ^{-1} \frac{63}{65}$

$=\cos ^{-1}\left(\frac{16}{65} \times \frac{63}{65}-\sqrt{1-\left(\frac{16}{65}\right)^{2}} \cdot \sqrt{1-\left(\frac{63}{65}\right)^{2}}\right)$

$=\cos ^{-1}\left(\frac{16}{65} \times \frac{63}{65}-\sqrt{\frac{65^{2}-16^{2}}{652}} \cdot \frac{65^{2}-63^{2}}{65^{2}}\right)$

$=\cos ^{-1}\left(\frac{1008}{65 \times 65}-\frac{\sqrt{(65+16)(65-16)}}{65} \cdot \frac{\sqrt{(65+63)(65-63)}}{65}\right)$

$=\cos ^{-1}\left(\frac{1008}{65 \times 65}-\frac{\sqrt{81 \times 49}}{65} \cdot \frac{\sqrt{128 \times 2}}{65}\right)$

$=\cos ^{-1}\left(\frac{1008}{65 \times 65}-\frac{7 \times 9}{65} \times \frac{16}{65}\right)$

$=\cos ^{-1}\left(\frac{1008}{65 \times 65}-\frac{1008}{65 \times 65}\right)$

$=\cos ^{-1}(0)$

$=\cos ^{-1} \cos \frac{\pi}{2}$

$=\frac{\pi}{2}$=R.H.S proved

(iii) $\sin \frac{-8}{17}+\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{77}{36}=\cos ^{-1} \frac{36}{85}$
Sol :
L.H.S
$=\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5} $

$=\sin ^{-1}\left(\frac{8}{17} \cdot \sqrt{1-\left(\frac{3}{5}\right)^{2}}+\frac{3}{5} \cdot \sqrt{1-\left(\frac{8}{17}\right)^{2}}\right)$

$=\sin ^{-1}\left(\frac{8}{17} \cdot \sqrt{\frac{5^{2}-3^{2}}{5^{2}}}+\frac{3}{5} \cdot\sqrt{\frac{17^{2}-8^{2}}{17^{2}}}\right) $

$=\sin ^{-1}\left(\frac{8}{17} \times \frac{4}{5}+\frac{3}{5} \times \frac{15}{17}\right)$

$=\sin ^{-1}\left(\frac{32}{85}+\frac{45}{85}\right)$

$=sin^{-1} \dfrac{77}{85}$

Let $=sin^{-1} \dfrac{77}{85}=\theta$ then $sin\theta= \dfrac{77}{85}$

$tan \theta=\dfrac{77}{36}$

∴$\theta=tan^{-1}\dfrac{77}{36}$

$=sin^{-1} \dfrac{77}{85}=tan^{-1}\dfrac{77}{36}$=R.H.S Proved

(iv) $\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{5}{13}=\sin ^{-1} \frac{56}{65}$ (Not correct)

(v) $\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}$

Sol :
L.H.S$=\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}$

$=\cos ^{-1}\left(\frac{4}{5} \times \frac{12}{13}-\sqrt{1-\frac{16}{25}} \cdot \sqrt{1-\frac{144}{169}}\right) $

$=\cos ^{-1}\left(\frac{48}{65}-\frac{3}{5} \times \frac{5}{13}\right)$

$=\cos ^{-1}\left(\frac{48}{65}-\frac{15}{65}\right) $

$=\cos ^{-1} \frac{33}{65}$=R.H.S proved

(vi) $\sin ^{-1} \frac{3}{5}-\sin ^{-1} \frac{8}{17}=\cos ^{-1} \frac{84}{85}$
Sol:
L.H.S $=\sin ^{-1} \frac{3}{5}-\sin ^{-1} \frac{8}{17}$

$=\sin ^{-1}\left(\frac{3}{5} \cdot \sqrt{1-\left(\frac{8}{17}\right)^{2}}-\frac{8}{17} \sqrt{1-\left(\frac{3}{5}\right)^{2}}\right)$

$=\sin ^{-1}\left(\frac{3}{5} \times \sqrt{\frac{17^{2}-8^{2}}{17^{2}}}-\frac{8}{17} \sqrt{\frac{5^{2}-3^{2}}{5^{2}}}\right)$

$=\sin ^{-1}\left(\frac{3}{5} \times \frac{\sqrt{289-64}}{17}-\frac{8}{17} \frac{\sqrt{25-9}}{5}\right)$

$=\sin ^{-1}\left(\frac{3}{5} \times \frac{15}{17}-\frac{8}{17} \times \frac{4}{5}\right)$

$=\sin ^{-1}\left(\frac{45}{85}-\frac{32}{85}\right)$

$=\sin ^{-1} \frac{13}{85}=\theta$ (Let)

then $\sin \theta=\frac{13}{89} \quad \therefore \cos \theta=\frac{84}{85}$

$\therefore \theta=\cos ^{-1} \frac{84}{85}$

$\sin ^{-1} \frac{13}{85}=\theta=\cos ^{-1} \frac{84}{85}$=R.H.S Proved

(vii) $\sin ^{-1} \frac{5}{13}+\cos \frac{1}{5}=\tan ^{-1} \frac{63}{16}$

Sol :

L.H.S $=\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}$

$=\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{4}{5}$

=$sin^{-1}\left(\dfrac{5}{13}.\sqrt{1-\left(\dfrac{4}{5}\right)^2}+\dfrac{4}{5}\sqrt{1-\left(\dfrac{5}{13}\right)^2}\right)$

$=\sin ^{-1}\left(\frac{5}{13} \cdot \frac{\sqrt{5^{2}-4^{2}}}{5}+\frac{4}{5} \cdot \frac{\sqrt{13^{2}-5^{2}}}{13}\right)$

$=\sin ^{-1}\left(\frac{5}{13} \cdot \frac{3}{5}+\frac{4}{5} \cdot \frac{12}{13}\right)$

$=\sin ^{-1}\left(\frac{15}{65}+\frac{48}{65}\right)$

$=\sin ^{-1}\left(\frac{63}{65}\right)=\theta$ then $\sin \theta=\frac{63}{65}$

∴$\tan \theta=\frac{63}{16} \Rightarrow \theta=\tan ^{-1} \frac{63}{16}$

∴$\sin ^{-1} \frac{63}{65}=\theta=\tan ^{-1} \frac{63}{16}$=R.H.S proved

(viii) $\sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}=\pi$
Sol :
LH.S $=\sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}$

$=\cos ^{-1} \frac{5}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}$

$=\cos ^{-1}\left(\frac{5}{13} \times \frac{4}{5}-\sqrt{1-\left(\frac{5}{19}\right)^{2}} \cdot \sqrt{1-\left(\frac{4}{5}\right)^{2}}\right)+\tan ^{-1} \frac{63}{16}$

$=\cos ^{-1}\left(\frac{20}{65}-\sqrt{\frac{13^{2}-5^{2}}{13^{2}}} \cdot \sqrt{\frac{5^{2}-4^{2}}{5^2}}\right)+\tan ^{-1} \frac{63}{16}$

$=\cos ^{-1}\left(\frac{20}{65}-\frac{\sqrt{169-25}}{13} \cdot \frac{\sqrt{25-16}}{5}\right)+\tan ^{-1} \frac{63}{16}$

$=\cos ^{-1}\left(\frac{20}{65}-\frac{12}{13} \times \frac{3}{5}\right)+\tan ^{-1} \frac{63}{16}$

$=\cos ^{-1}\left(\frac{20}{65}-\frac{36}{65}\right)+\tan ^{-1} \frac{63}{16}$

$=\cos ^{-1}\left(-\frac{16}{65}\right)+\tan ^{-1} \frac{63}{16}$

$=\pi-\cos ^{-1} \frac{16}{65}+\tan ^{-1} \frac{63}{16}$

$=\pi-\tan ^{-1} \frac{63}{16}+\tan ^{-1} \frac{63}{16}$

$=\pi$ =R.H.S proved

(ix) $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$

Sol :

l.H.S$=2 \sin ^{-1} \frac{3}{5}$

$=\sin ^{-1}\left(2 \cdot\dfrac{ 3}{5} \cdot \sqrt{1-\left(\dfrac{3}{5}\right)^{2}}\right)$

[∵ 2sin^-1x=sin^-1(2x.√1-x²)]

$=sin^{-1} \left(\dfrac{6}{5}.\sqrt{\dfrac{5^2-3^2}{5^2}}\right)$

$=sin^{-1} \left(\dfrac{6}{5}\times \dfrac{4}{5}\right)$

$=sin^{-1}\left(\dfrac{24}{25}\right)$

$=tan^{-1} \dfrac{24}{7}$=R.H.S proved

(x) $\dfrac{9\pi}{8}-\dfrac{9}{4}sin^{-1}\dfrac{1}{3}=\dfrac{9}{4}sin^{-1}\dfrac{2\sqrt{2}}{3}$

Sol :
$\frac{9 \pi}{8}=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}+\frac{9}{4} \sin ^{-1} \frac{1}{3}$

$\frac{9 \pi}{8}=\frac{9}{4}\left(\sin ^{-1} \frac{2 \sqrt{2}}{3}+\sin ^{-1} \frac{1}{3}\right)$

Now , R.H.S=$=\frac{9}{4}\left(\sin ^{-1} \frac{2 \sqrt{2}}{3}+\sin ^{-1} \frac{1}{3}\right)$

$=\dfrac{9}{4}\left[\sin ^{-1}\left(\dfrac{2 \sqrt{2}}{3}\sqrt{1-\dfrac{1}{9}}+\dfrac{1}{3} \cdot \sqrt{1-\dfrac{8}{9}}\right)\right]$

$=\frac{9}{4}\left[\sin ^{-1}\left(\frac{2 \sqrt{2}}{3} \cdot \frac{\sqrt{9-1}}{3}+\frac{1}{3} \cdot \frac{\sqrt{9-8}}{3}\right)\right]$

$=\frac{9}{4}\left[\sin ^{-1}\left(\frac{2 \sqrt{2}}{3} \cdot \frac{2 \sqrt{2}}{3}+\frac{1}{3} \cdot \frac{1}{3}\right)\right]$

$=\frac{9}{4}\left[\sin ^{-1}\left(\frac{8}{9}+\frac{1}{9}\right)\right]$

$=\frac{9}{4}\left[\sin ^{-1}\left(\frac{9}{9}\right)\right]$

=$\dfrac{9}{4}\left[sin^{-1}(1)\right]$

=$\dfrac{9}{4}sin^{-1} sin\dfrac{\pi}{2}$

=$\dfrac{9}{4}\times \dfrac{\pi}{2}=\dfrac{9\pi}{8}$=L.H.S Proved

(xii) $\sin ^{-1} x+\sin ^{-1} y=\cos ^{-1}(\sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}-x y),$ when
$x \in[0,1], y \in[0,1]$

Sol :

L.H.S=sin^-1x+sin^-1y

=$cos^{-1}\sqrt{1-x^2}+cos^{-1}\sqrt{1-y^2}$

=$cos^{-1}\left(\sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}-\sqrt{1-(\sqrt{1-x^{2}})^{2}} \cdot \sqrt{1-(\sqrt{1-y^{2}})^{2}}\right)$

=$\cos ^{-1}(\sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}-\sqrt{1-\left(1-x^{2}\right)} \cdot \sqrt{1-\left(1-y^{2}\right)})$

=$\cos ^{-1}(\sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}-\sqrt{1-1+x^{2}} \cdot \sqrt{1-1+y^{2}})$

=$\cos ^{-1}(\sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}-\sqrt{x^{2}} \cdot \sqrt{y^{2}})$

=$\cos ^{-1}(\sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}-x y)$=R.H.S Proved

Question 9

Prove that 
(i) $4\left(sin^{-1}\dfrac{1}{\sqrt{10}+cos^{-1}\dfrac{2}{\sqrt{5}}}\right)=\pi$
Sol :
L.H.S=$=4\left(\sin ^{-1} \frac{1}{\sqrt{10}}+\cos ^{-1} \frac{2}{\sqrt{5}}\right)$

$=4\left(\cos ^{-1} \frac{3}{\sqrt{10}}+\cos ^{-1} \frac{2}{\sqrt{5}}\right)$

$=4\left[\cos ^{-1}\left(\frac{3}{\sqrt{10}} \times \frac{2}{\sqrt{5}}-\sqrt{1-\frac{9}{10}} \cdot \sqrt{1-\frac{4}{5}}\right)\right]$

$=4\left[\cos ^{-1}\left(\frac{6}{\sqrt{50}}-\sqrt{\frac{10-9}{10}} \cdot \sqrt{\frac{5-4}{5}}\right)\right]$

$=4\left[\cos ^{-1}\left(\frac{6}{\sqrt{50}}-\frac{1}{110} \cdot \frac{1}{\sqrt{5}}\right)\right]$

$=4\cos ^{-1}\left(\frac{6}{\sqrt{50}}-\frac{1}{\sqrt{50}}\right)$

$=scos^{-1}\dfrac{6-1}{\sqrt{50}}$

$=4cos^{-1}\dfrac{5}{\sqrt{25}\times \sqrt{2}}$

$=4cos^{-1}\dfrac{5}{5\times \sqrt{2}}$

$=4cos^{-1} \dfrac{1}{\sqrt{2}}=4cos^{-1}cos\dfrac{\pi}{4}$

$4 \times \dfrac{\pi}{4}=\pi$=R.H.S Proved

(ii) cos(2 sin^-1x)=1-2x²

Sol :

L.H.S=cos(2 sin^-1x)

Let 2 sin^-1x =θ

$\sin ^{-1}(2 x \cdot \sqrt{1-x^{2}})=\theta$

$\sin \theta=2 x \cdot \sqrt{1-x^{2}}$

∴$\cos \theta=\sqrt{1-\sin ^{2} \theta}=\sqrt{1-4 x^{2}\left(1-x^{2}\right)}$

$\cos \theta=\sqrt{1-4 x^{2}+4 x^{4}}=\sqrt{4 x^{4}+1-4 x^{2}}$

$\cos \theta=\sqrt{\left(1-2 x^{2}\right)^{2}}=\left(1-2 x^{2}\right)$

θ=cos^-1(1-2x²)

∴ 2sin^-1x=θ=cos^-1(1-2x²)

Now L.H.S=cos(2sin^-1x)

=cos.cos^-1(1-2x²)

=1-2x²=R.H.S Proved

(iii) cos(sec^-1x+cosec^-1x)=0, |x|≥1

Sol :

L.H.S=cos(sec^-1x+cosec^-1x)

=$cos \dfrac{\pi}{2}=0$=R.H.S Proved

(iv) $\frac{1}{2} \cos ^{-1} x=\sin ^{-1} \sqrt{\frac{1-x}{2}}=\cos ^{-1} \frac{\sqrt{1+x}}{2}=\tan ^{-1} \frac{\sqrt{1-x^{2}}}{1+x}$

Sol :

L.H.S=$\dfrac{1}{2}cos^{-1}x=\theta$ then $cos^{-1}x=2\theta$

⇒x=cos2θ

⇒x=1-2sin²θ

⇒2sin²θ=1-x

⇒$sin2\theta=\dfrac{1-x}{2}$

⇒$sin\theta =\sqrt{\dfrac{1-x}{2}}$

∴$\dfrac{1}{2}cos^{-1}x=\theta=sin\sqrt{\dfrac{1-x}{2}}$=R.H.S Proved

again , L.H.S$=sin^{-1}\sqrt{\dfrac{1-x}{2}}=\theta$ then $sin\theta =\sqrt{\dfrac{1-x}{2}}$

∴$\cos \theta=\frac{1-\sin ^{2} \theta}{2}=\sqrt{1-\frac{1-x}{2}}=\sqrt{\frac{2-1+x}{2}}$

$\cos \theta=\sqrt{\frac{1+x}{2}}$

∴$\theta=cos^{-1}\sqrt{\dfrac{1+x}{2}}$

Now L.H.S=$\sin ^{-1} \sqrt{\frac{1-x}{2}}=\theta=\cos ^{-1} \sqrt{\frac{1+x}{2}}$=R.H.S Proved

Question 10

Prove that

(i) $\sin ^{-1} x+\cos ^{-1} y=\tan ^{-1}\left(\frac{x y+\sqrt{\left(1-x^{2}\right)\left(1-y^{2}\right)}}{y \sqrt{\left(1-x^{2}\right)}-x \sqrt{\left(1-y^{2}\right)})}\right.$

Sol :

L.H.S$=\sin ^{-1} x+\cos ^{-1} y $

$=\cos ^{-1} \sqrt{1-x^{2}}+\cos ^{-1} y $

$=\cos ^{-1}(\sqrt{1-x^{2}} \cdot y-\sqrt{1-\left(1-x^{2}\right)} \cdot \sqrt{1-y^{2}}) $

$=\cos ^{-1}(y \cdot \sqrt{1-x^{2}}-\sqrt{x^{2}} \cdot \sqrt{1-y^{2}})$

=$=\cos \theta=(y \sqrt{1-x^{2}}-x \sqrt{1-y^{2}})$

$\sin \theta=\sqrt{1-\cos ^{2} \theta}$

$\sin \theta=\sqrt{1-(y \sqrt{1-x^{2}}-x \sqrt{1-y^{2}})^{2}}$

$\sin \theta=\sqrt{1-\left[y^{2}\left(1-x^{2}\right)+x^{2}\left(1-y^{2}\right)-2 x y \sqrt{1-x^{2}}\right. \sqrt{1-y^2}}$

$\sin \theta=\sqrt{1-\left(y^{2}-x^{2} y^{2}+x^{2}-x^{2} y^{2}-2 x y \sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}\right)}$

$\sin \theta=\sqrt{1-y^{2}+x^{2} y^{2}-x^{2}+x^{2} y^{2}+2 x y \sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}}$

$\sin \theta=\sqrt{1-x^{2}-y^{2}+2 x^{2} y^{2}+2 x y \sqrt{1-x^{2}} \cdot \sqrt{1-y^{2}}}$

$\sin \theta=\sqrt{\left(x y+\sqrt{\left(1-x^{2}\right)} \cdot \sqrt{\left(1-y^{2}\right)}\right)^2}$

$\sin \theta=x y+\sqrt{\left(1-x^{2}\right)} \cdot \sqrt{\left(1-y^{2}\right)}$

∴$\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{x y+\sqrt{\left(1-x^{2}\right)} \cdot \sqrt{\left(1-y^{2}\right)}}{y \cdot \sqrt{\left(1-x^{2}\right)}-x \sqrt{\left(1-y^{2}\right)}}$

∴$\theta=\tan ^{-1}\left(\frac{x y+\sqrt{\left(1-x^{2}\right)} \cdot \sqrt{\left(1-y^{2}\right)}}{y \cdot \sqrt{\left(1-x^{2}\right)-x} \sqrt{\left(1-y^{2}\right)})}\right.$

Now, $\cos ^{-1}(y \sqrt{1-x^{2}}-x \sqrt{1-y^{2}})=\theta$

∴$\sin ^{-1} x+\cos ^{-1} y=\tan ^{-1}\left(\frac{x y+\sqrt{\left(1-x^{2}\right)} \cdot \sqrt{\left(1-y^{2}\right)}}{y \cdot \sqrt{\left(1-x^{2}\right)-x \sqrt{1-y^{2}}}}\right)$

Proved

(ii) $\tan ^{-1} x+\tan ^{-1} y=\frac{1}{2} \sin ^{-1} \frac{2(x+y)(1-x y)}{\left(1+x^{2}\right)\left(1+y^{2}\right)}$
Sol :

Given $\tan ^{-1} x+\tan ^{-1} y$ $=\frac{1}{2} \sin ^{-1} \frac{2(x+y)(1-x y)}{\left(1+x^{2}\right)\left(1+y^{2}\right)}$

$=2\left(\tan ^{-1} x+\tan ^{-1} y\right)=\sin ^{-1} \frac{2(x+y)(1-x y)}{\left(1+x^{2}\right)\left(1+y^{2}\right)}$

Now, L.H.S $=2\left(\tan ^{-1} x+\tan ^{-1} y\right)$

$=2 \tan ^{-1}\left(\frac{x+y}{1-x y}\right)$

$\therefore 2tan^{-1}x=sin^{-1}\dfrac{2x}{1+x^2}$

=$sin^{-1}\dfrac{2\left(\frac{x+y}{1-xy}\right)}{1+\left(\frac{x+y}{1-xy}\right)^2}$

=$\sin ^{-1} \dfrac{\frac{2(x+y)}{(1-x y)}}{1+\frac{(x+y)^2}{(1-xy)^2}}$

=$\sin ^{-1}\dfrac{\frac{ 2(x+y)}{(1-x y)} \times(1-x y)^{2}}{(1-xy)^2+(x+y)^2}$

=$\sin ^{-1} \dfrac{2(x+y)(1-x y)}{1+x^{2} y^{2}-2 x y+x^{2}+y^{2}+2 x y}$

=$\sin ^{-1} \dfrac{2(x+y)(1-x y)}{1+x^{2}+x^{2} y^{2}+y^{2}}$

=$\sin ^{-1} \dfrac{2(x+y)(1-x y)}{\left(1+x^{2}\right)+y^{2}\left(1+x^{2}\right)}$

=$\sin ^{-1} \dfrac{2(x+y)(1-x y)}{\left(1+x^{2}\right)\left(1+y^{2}\right)}$

=R.H.S Proved

Question 11

Prove that:
(i) 2tan^-1(cosec.tan^-1x-tan.cot^-1x)=tan^-1x
Sol :
L.H.S=2tan^-1(cosec.tan^-1x-tan.cot^-1x)

$=2 \tan ^{-1}\left(\operatorname{cosec} \tan ^{-1} x-\tan \cdot \tan ^{-1} \frac{1}{x}\right)$

$=2 \tan ^{-1}\left(\operatorname{cosec} \cdot \operatorname{cosec}^{-1} \frac{\sqrt{1+x^{2}}}{x}-\tan \tan \frac{-1}{x}\right)$

=$2 \tan ^{-1}\left(\frac{\sqrt{1+x^{2}}}{x}-\frac{1}{x}\right)$

$=2 \tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$

$=\tan ^{-1} \frac{2\left(\frac{\sqrt{1+x^{2}-1}}{x}\right)}{1-\left(\frac{\sqrt{1+x^{2}-1}}{x}\right)^{2}}$

$=\tan ^{-1} \frac{2(\sqrt{1+x^{2}}-1)}{\frac{x^{2}-(\sqrt{1+x^{2}-1})^{2}}{x^{2}}}$

=$\tan ^{-1} \frac{2(\sqrt{1+x^{2}}-1) x}{x^{2}-\left(1+x^{2}+1-2 \sqrt{1+x^{2}}\right)}$

$=\tan ^{-1} \frac{2(\sqrt{1+x^{2}}-1) \cdot x}{x^{2}-1-x^{2}-1+2 \sqrt{1+x^{2}}}$

$=\tan ^{-1} \frac{2(\sqrt{1+x^{2}-1}) \cdot x}{2 \sqrt{1+x^{2}-2}}$

$=\tan ^{-1} \frac{2(\sqrt{1+x^{2}}-1) \cdot x}{2(\sqrt{1+x^{2}-1})}$

$=\tan ^{-1} x$=R.H.S proved

(ii) $\cos. \tan ^{-1} .\sin .\cot ^{-1} x=\sqrt{\frac{x^{2}+1}{x^{2}+2}}$
Sol :
L.H.S $=\cos. \tan ^{-1} .\sin. \cot ^{-1} x$

$=\cos. \tan ^{-1}. \sin. \sin ^{-1} \frac{1}{\sqrt{1+x^{2}}}$ $\left(\because cot^{-1}=sin^{-1}\dfrac{1}{\sqrt{1+x^2}}\right)$

=$cos.tan^{-1}\dfrac{1}{\sqrt{1+x^2}}$ $\left(tan^{-1}\dfrac{1}{\sqrt{1+x^2}}=cos^{-1}\dfrac{\sqrt{1+x^2}}{x^2+2}\right)$

=$cos.cos^{-1}\dfrac{\sqrt{x^2+1}}{\sqrt{x^2+2}}$

=$\dfrac{\sqrt{x^2+1}}{\sqrt{x^2+2}}$=R.H.S proved

TYPE-IV

Question 12
(i) $tan^{-1}x+tan^{-1}y+tan^{-1}z=\dfrac{\pi}{2}$
Prove that yz+zx+xy=1
Sol:

=$\tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{2}-\tan ^{2} z$

$=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\cot ^{-1} z$ $\left(\because \tan ^{-1} z+\cot \frac{-1}{z}=\frac{\pi}{2}\right)$

$=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\tan ^{-1} \frac{1}{z}$

=$\tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\cot ^{-1} z \quad\left(\because \tan ^{-1} z+\cot ^{-1}z=\frac{\pi}{2}\right)$

=$\tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\tan ^{-1} \frac{1}{z}$

$=\frac{x+y}{1-x y}=\frac{1}{z} \Rightarrow z(x+y)=1-x y$

$=z x+z y=1-x y$

$=y z+z x+x y=1$ proved

(ii) If $\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\pi,$ prove that

x+y+z = x y z
Sol :
Given $\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\pi$

let $\tan ^{-1} \alpha$ then $\tan \alpha=x$
$\tan ^{-1} y=\beta$ them $\tan \beta=y$

$\tan ^{-1} z=\gamma $ then $\tan \gamma=z$

Now, $\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\pi$

$=\tan ^{-1} x+\tan ^{-1} y=\pi-\tan ^{-1} z$

$=\gamma+\beta=\pi-y$

$=\tan (\alpha+\beta)=\tan (\pi-y)$

$\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}=-\tan y$

$=\frac{x+y}{1-x y}=-\frac{z}{1} \Rightarrow x+y=-z(1-x y)$

$=x+y=-z+x y z=x+y+z=x y z$ proved

Question 13

(i) If $\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2}$, prove that

$x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=1$

Sol :

Given $\sin ^{-1} x+\sin ^{-1} y=\frac{x}{2}$

$=\sin ^{-1}(x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}})=\frac{x}{2}$

$=x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=\sin \frac{x}{2}$

$=x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=1$ proved

(ii) If $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=x$, prove that
$x \sqrt{1-x^{2}}+y \sqrt{1-y^{2}}+z \sqrt{1-z^{2}}=2 x y z$

Sol :

Given $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=x$

let $\sin ^{-1} x=A$ then $\sin A=x $ $ \therefore \cos A=\sqrt{1-x^{2}}$

$\sin ^{-1} y=B$ then $\sin B=y $ $ \therefore \cos B=\sqrt{1-y^{2}}$

$\sin ^{-1} z=c \quad$ then $\quad \sin C=z \quad \therefore \cos C=\sqrt{1-z^{2}}$

Now, A+B+C=π

∴sin 2A+sin 2B+sin 2C=4 sin A.sin B.sin C

2sinA.cosA+2sinB.cosB+2sinC.cosC=4sinA.sinB.sinC

=2(sinA.cosA+sinB.cosB+sinC.cosC)=4sinA.sinB.sinC

=sinA.cosA+sinB.cosB+sinC.cosC=2sinA.sinB.sinC

=$x \cdot \sqrt{1-x^{2}}+y \cdot \sqrt{1-y^{2}}+z \cdot \sqrt{1-z^{2}}=2 \cdot x y z$ Proved

(iii) Establish the algebraic relation between x, y, z if $\tan ^{-1} x, \tan ^{-1} y, \tan ^{-1} z$ are in A.P and if further x, y, z are also in A.P. them prove that x=y=z

Sol:

Given $\tan ^{-1} x, \tan ^{-1} y, \tan ^{-1} z$ are in A.P

x,y,z are also in A.P

∴x-y=y-z ⇒x+z=2y...(i)

also , $\tan ^{-1} x+\tan ^{-1} z=2 \tan ^{-1} y$

$=\tan ^{-1}\left(\frac{x+z}{1-x z}\right)=\tan ^{-1} \frac{2 y}{1-y^{2}}$

$=\frac{x+z}{1-x z}=\frac{2 y}{1-y^{2}}$

$=\frac{2 y}{1-x z}=\frac{2 y}{1-y^{2}}=\frac{1}{1-x z}=\frac{1}{1-y^{2}}$

$=1-y^{2}=1-x z=-y^{2}=-x z$

$=y^{2}=x z$..(ii)

From equation (i) $\left(\dfrac{x+z}{2}\right)=y$

∴$y^2=\dfrac{(x+z)^2}{4}$...(iii)

Comparing equation (ii) and (iii) we get

xz=$\dfrac{(x+z)^2}{4}$⇒4xz=x²+z²+2xz

=x²+z²+2xz=0 ⇒(x-z)²=0

⇒(x-z)=0

⇒x=z..(iv)

Now $\frac{x+z}{1-x z}=\frac{2 y}{1-y^{2}} \Rightarrow \frac{2 y}{1-x^{2}}=\frac{2 y}{1-y^{2}}$

=1-x²=1-y²=x²=y²

x=y ..(v)

From equation (iv) and (v) we get

x=y=z Proved

Question 14

(i) $\cot ^{-1} x+\sin ^{-1} \frac{1}{\sqrt{5}}=\frac{\pi}{4}$

Sol :

$\Rightarrow \quad \tan ^{-1} \frac{1}{x}+\tan ^{-1} \frac{1}{2}=\frac{\pi}{4}$ $\left[\because \sin ^{-1} \frac{1}{\sqrt{5}}=\tan ^{-1} \frac{1}{2}\right]$

$\Rightarrow \frac{\left(\frac{1}{x}+\frac{1}{2}\right)}{\left(1-\frac{1}{x} \cdot \frac{1}{2}\right)}=\frac{x}{4}$

$\frac{\frac{2+x}{2 x}}{\frac{2x-1}{2 x}}=\tan \frac{\pi}{4}$

$\Rightarrow \frac{2+x}{2 x-1}=1 \Rightarrow 2+x=2 x-1$

=2x-x=3

⇒x=3

(ii) $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$
Sol :
Given $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$

$\Rightarrow \quad \sin ^{-1} \frac{1}{5}+\cos ^{-1} x=\sin ^{-1}(1)$

$\Rightarrow \quad \sin ^{-1} \frac{1}{5}+\cos ^{-1} x=\sin ^{-1} \sin \frac{\pi}{2}$

$\Rightarrow \quad \sin ^{-1} \frac{1}{5}+\cos ^{-1} x=\frac{\pi}{2}$

$\Rightarrow \quad \sin ^{-1} \frac{1}{5}=\frac{\pi}{2}-\cos ^{-1} x$

$\Rightarrow \sin \frac{-1}{5}=\sin ^{-1} x$

$\therefore x=\frac{1}{5} \quad$ Ans

Question 15

Solve 

(i) $\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$

Sol :

$\Rightarrow \tan ^{-1}\left(\frac{2 x+3 x}{1-2 x \cdot 3 x}\right)=\frac{\pi}{4}$

$\Rightarrow \frac{5 x}{1-6 x^{2}}=\tan \frac{\pi}{4} \Rightarrow \frac{5 x}{1-6 x^{2}}=1$

$\Rightarrow \quad 5 x=1-6 x^{2} \quad \Rightarrow \quad 6 x^{2}+5 x-1=0$

$\Rightarrow \quad 6 x^{2}+6 x-x-1=0$

$\Rightarrow \quad 6 x(x+1)-1(x+1)=0$

$\Rightarrow(x+1)(6 x-1)=0$

$x=-1, \quad x=\frac{1}{6} \quad$ Ans

(ii) $\tan ^{-1} x+\tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{\pi}{3}$

Sol :

$\Rightarrow \tan ^{-1} x+2 \tan ^{-1} x=\frac{\pi}{3}$

$\Rightarrow 3 \tan ^{-1} x=\frac{\pi}{3}$

$\Rightarrow \quad \tan ^{-1} x=\frac{\pi}{9}$

$\Rightarrow \quad x=\tan \frac{\pi}{9}$

(iii) $\tan ^{-1} \frac{1}{2}=\cot ^{-1} x+\tan ^{-1} \frac{1}{7}$

Sol :

$\tan ^{-1} \frac{1}{2}-\tan ^{-1} \frac{1}{7}=\tan^{-1} \frac{1}{x}$

⇒$\tan ^{-1} \frac{\left(\frac{1}{2}-\frac{1}{7}\right)}{\left(1+\frac{1}{2} \cdot \frac{1}{7}\right)}=\tan \frac{-1}{x}$

⇒$\tan ^{-1}\left(\frac{7-2}{14+1}\right)=\tan \frac{1}{x} \Rightarrow \frac{5}{15}=\frac{1}{x} \Rightarrow x=3$

Question 16

(i) $\tan ^{-1}(x-1)+\tan ^{-1} x+\tan ^{-1}(x+1)=\tan ^{-1} 3 x$
Sol :
$\Rightarrow \quad \tan ^{-1}(x-1)+\tan ^{-1}(x+1)=\tan ^{-1} 3 x-\tan ^{-1} x$

$\Rightarrow \tan ^{-1}\left(\frac{(x-1)+(x+1)}{1-(x-1)(x+1)}\right)=\tan ^{-1}\left(\frac{3 x-x}{1+3 x \cdot x}\right)$

$\Rightarrow \quad \frac{2 x}{1-\left(x^{2}-1\right)}=\frac{2 x}{1+3 x^{2}} \Rightarrow \frac{2 x}{2-x^{2}}=\frac{2 x}{1+3 x^{2}}$

$\begin{aligned} \Rightarrow 1+3 x^{2}=2-x^{2} & \Rightarrow 4 x^{2}=1 \\ & \Rightarrow x^{2}=\frac{1}{4} \end{aligned}$

$\Rightarrow x=\pm \dfrac{1}{2}$

(ii) $\tan ^{-1} \frac{x+1}{x-1}+\tan ^{-1} \frac{x-1}{x}=\pi+\tan ^{-1}(-7)$
Sol :
$\tan ^{-1} \frac{\left(\frac{x+1}{x-1}+\frac{x-1}{x}\right)}{1-\left(\frac{x+1}{x-1}\right)\left(\frac{x-1}{x}\right)}=\pi-\tan ^{-1} 7$

$\Rightarrow \tan ^{-1} \dfrac{\left(\frac{x(x+1)+(x-1)(x-1)}{x(x-1)}\right)}{\frac{x(x-1)-(x+1)(x-1)}{x(x-1)}}=\pi-\tan 7$

$\Rightarrow \tan ^{-1} \frac{\left[x(x+1)+(x-1)^{2}\right]}{x(x-1)-\left(x^{2}-1\right)}=\pi-\tan ^{-1} 7$

$\Rightarrow \frac{x(x+1)+(x-1)^{2}}{x(x-1)-\left(x^{2}-1\right)}=\tan \left(\pi-\tan ^{-1} 7\right)$

$\Rightarrow \frac{x^{2}+x+x^{2}+1-2 x}{x^{2}-x-x^{2}+1}=-\tan .\tan ^{-1} 7$

$\Rightarrow \frac{2 x^{2}-x+1}{1-x}=-7$ $(\because \tan (x-\theta)=-\tan\theta$

$\Rightarrow \quad 2 x^{2}-x+1=-7(1-x)$
$\Rightarrow 2 x^{2}-x+1=-7+7 x$
$\Rightarrow \quad 2 x^{2}-8 x+8=0$
$\Rightarrow \quad x^{2}-4 x+4=0$
$\Rightarrow x^{2}-2 x-2 x+4=0$
⇒x(x-2)-2(x-2)=0
⇒(x-2)(x-2)=0
⇒∴ x=2

(iii) $\tan ^{-1} \frac{x-1}{x-2}+\tan ^{-1} \frac{x+1}{x+2}=\frac{\pi}{4}$

Sol :

$\tan ^{-1} \frac{\left(\frac{x-1}{x-2}+\frac{x+1}{x+2}\right)}{1-\left(\frac{x-1}{x-2}\right)\left(\frac{x+1}{x+2}\right)}=\frac{\pi}{4}$

$\Rightarrow \frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)-(x-1)(x+1)}=\tan \frac{\pi}{4}$

$\Rightarrow \frac{x^{2}+2 x-x-2+x^{2}-2 x+x-2}{x^{2}-4-\left(x^{2}-1\right)}=1$

$\Rightarrow \frac{2 x^{2}-4}{x^{2}-4-x^{2}+1} \Rightarrow \frac{2 x^{2}-4}{-3}=1$

$\Rightarrow \quad 2 x^{2}-4=-3 \quad \Rightarrow \quad 2 x^{2}=-3+4$

$\Rightarrow \quad 2 x^{2}=1 \quad \Rightarrow \quad x^{2}=\frac{1}{2}$

∴ $x=\pm \dfrac{1}{\sqrt{2}}$

Question 17

Solve

$\sin ^{-1} \frac{2 \alpha}{1+\alpha^{2}}+\sin ^{-1} \frac{2 \beta}{1+\beta^{2}}=2 \tan ^{-1} x|\alpha| \leq|\beta| \leq 1$

Sol :

$\Rightarrow \quad 2 \tan ^{-1} \alpha+2 \tan ^{-1} \beta=2 \tan ^{-1} x$

$\Rightarrow \quad 2\left(\tan ^{-1} \alpha+\tan ^{-1} \beta\right)=2 \tan ^{-1} x$

$\tan ^{-1} \alpha+\tan ^{-1} \beta=\tan ^{-1} x$

$\Rightarrow \quad \tan ^{-1}\left(\frac{\alpha+\beta}{1-\alpha \cdot \beta}\right)=\tan ^{-1} x$

$\Rightarrow \quad x=\frac{a+\beta}{1-\alpha \beta} \quad$ Ans

Question 18

Solve

(i) $\tan ^{-1} a x+\frac{1}{2} \sec ^{-1} b x=\frac{\pi}{4}$

Sol :

$\Rightarrow 2 \tan ^{-1}a x+\sec ^{-1} b x=\frac{\pi}{2}$

$\Rightarrow \tan ^{-1} \frac{2 a x}{1-a^{2} x^{2}}=\frac{x}{2}-\sec ^{-1} b x$

$\Rightarrow \frac{2 a x}{1-a^{2} x^{2}}=\tan \left(\frac{\pi}{2}-\sec ^{-1} b x\right)$

$\left[\because \tan \left(\frac{x}{2}-a\right)=\cot \theta \right]$

$\Rightarrow \frac{2 a x}{1-a^{2} x^{2}}=\tan \left(\frac{x}{2}-\cot ^{-1} \frac{1}{\sqrt{b^{2} x^{2}-1}}\right)$ 

$\Rightarrow \frac{2 a x}{1-a^{2} x^{2}}=\cot \cot ^{-1} \frac{1}{\sqrt{b^{2} x^{2}-1}}$

$\Rightarrow \frac{2 a x}{1-a^{2} x^{2}}=\frac{1}{\sqrt{b^{2} x^{2}-1}}$

$\Rightarrow 2ax\sqrt{b^{2} x^{2}-1}=\left(1-a^{2} x^{2}\right)$

$\Rightarrow 4 a^{2} x^{2}\left(b^{2} x^{2}-1\right)=\left(1-a^{2} x^{2}\right)$

$\Rightarrow 4 a^{2} b^{2} x^{4}-4 a^{2} x^{2}=1+a^{4} x^{4}-2 a^{2} x^{2}$

$\Rightarrow 4 a^{2} b^{2} x^{4}=1+a^{4} x^{4}-2 a^{2} x^{2}+4 a^{2} x^{2}$

$\Rightarrow 4 a^{2} b^{2} x^{4}=1+a^{4} x^{4}+2 a^{2} x^{2}$

$\Rightarrow \left(1+a^{2} x^{2}\right)^{2}=4 a^{2} b^{2} x^{4}$

$\Rightarrow 1+a^{2} x^{2}=\sqrt{4 a^{2}-b^{2} x^{4}}=2 a b x^{2}$

$\Rightarrow \quad 2 a b x^{2}-a^{2} x^{2}=1$

$\Rightarrow \quad x^{2} \times\left(2 a b-a^{2}\right)=1$

$\Rightarrow \quad x^{2}=\frac{1}{\left(2 a b-a^{2}\right)}$

$\Rightarrow \quad x=\pm \frac{1}{\sqrt{2 a b-a^{2}}} $ Ans

(iii) $\tan \left(\cos ^{-1} x\right)=\sin \left(\tan ^{-1} 2\right)$
Sol :
$\Rightarrow \tan \left(\tan ^{-1} \frac{\sqrt{1-x^{2}}}{x}\right)=\sin \left(\sin ^{-1} \frac{2}{\sqrt{5}}\right)$

$\Rightarrow \quad \frac{\sqrt{1-x^{2}}}{x}=\frac{2}{\sqrt{5}} \Rightarrow \frac{\left(1-x^{2}\right)}{x^{2}}=\frac{4}{5}$

$\Rightarrow \quad 5\left(1-x^{2}\right)=4 x^{2} \Rightarrow 5-5 x^{2}=4 x^{2}$

$\Rightarrow \quad 9 x^{2}=5$

$\Rightarrow \quad x^{2}=\frac{5}{9}$

$\Rightarrow \quad x=\sqrt{\frac{5}{9}}=\pm \frac{\sqrt{5}}{3}$

(iii) $\tan \left(\sec ^{-1} \frac{1}{x}\right)=\sin \cos ^{-1} \frac{1}{\sqrt{5}}$

Sol :

$\Rightarrow \tan \left(\cos ^{-1} x\right)=\sin \left(\sin ^{-1} \frac{2}{\sqrt{5}}\right)$

$\Rightarrow \tan \left(\tan \frac{\sqrt{1-x^{2}}}{x}\right)=\sin \left(\sin ^{-1} \frac{2}{\sqrt{5}}\right)$

$\Rightarrow \quad \frac{\sqrt{1-x^{2}}}{x}=\frac{2}{\sqrt{5}} \Rightarrow \frac{\left(1-x^{2}\right)}{x^{2}}=\frac{4}{5}$

$\Rightarrow \quad 5\left(1-x^{2}\right)=4 x^{2} \Rightarrow 5-5 x^{2}=4 x^{2}$

$\Rightarrow 4 x^{2}+5 x^{2}=5 \Rightarrow 9 x^{2}=5$

$\Rightarrow \quad x^{2}=\frac{5}{9}$

$\therefore x=\pm \frac{\sqrt{5}}{3} $ Ans

(iv) $\cos \left(\tan ^{-1} x\right)=\sin \left(\cot ^{-1} \frac{3}{4}\right)$
Sol :
$\Rightarrow \cos \left(\cos ^{1} \frac{1}{\sqrt{1+x^{2}}}\right)=\sin \left(\sin ^{-1} \frac{4}{5}\right)$

$\Rightarrow \frac{1}{\sqrt{1+x^{2}}}=\frac{4}{5} \Rightarrow \frac{1}{1+x^{2}}=\frac{16}{25}$

$\Rightarrow \quad 16\left(1+x^{2}\right)=25$

$\Rightarrow \quad 16+16 x^{2}=25$

$\Rightarrow \quad 16 x^{2}=25-16=9$

$16 x^{2}=9$

$\Rightarrow \quad x^{2}=\frac{9}{16}$

$\Rightarrow \quad x=\pm \frac{3}{4}$

Question 19

If $tan^{-1}\left(\frac{x-2}{x-4}\right)+\tan ^{-1}\left(\frac{x+2}{x+4}\right)=\frac{x}{4},$ Find the value of x

Sol :

$\Rightarrow \tan ^{-1}\left(\frac{\left.\frac{x-2}{x-4}+\frac{x+2}{x+4}\right)}{x-\left(\frac{x-2}{x-4}\right)\left(\frac{x+2}{x+4}\right)}\right)=\frac{x}{4}$

$\Rightarrow \tan ^{-1} \frac{\left(\frac{(x-2)(x+4)+(x+2)(x-4)}{(x-4)(x+4)}\right)}{\left(\frac{(x-4)(x+4)-(x-2)(x+2)}{(x-4)(x+4)}\right)}=\frac{\pi}{4}$

$\Rightarrow \quad \frac{(x-2)(x+4)+(x+2)(x-4)}{(x-4)(x+4)-(x-2)(x+2)}=\tan \frac{\pi}{4}$

$\Rightarrow \frac{x^{2}+4 x-2 x-8+x^{2}-4 x+2 x-8}{\left(x^{2}-16\right)-\left(x^{2}-4\right)}=1$

$\Rightarrow \quad \frac{2 x^{2}-16}{x^{2}-16-x^{2}+4}=1$

⇒$\dfrac{2x^2-16}{-12}=1$

⇒2x²-12+16⇒2x²=4

⇒x²=2 $\Rightarrow x=\pm \sqrt{2}$