Class 11: Maths Chapter 3 solutions. Complete Class 11 Maths Chapter 3 Notes.
Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 1): Chapter 3- Trigonometry – II
Maharashtra Board 11th Maths Chapter 3, Class 11 Maths Chapter 3 solutions
Ex 3.1
∴ tan 8θ – tan8θ.tan5θ.tan3θ = tan5θ + tan 3θ
∴ tan 8θ – tan 5θ – tan 3θ = tan 8θ.tan 5θ.tan 3θ
Ex 3.2
Question 1.
Find the values of:
i. sin 690°
ii. sin 495°
iii. cos 315°
iv. cos 600°
v. tan 225°
vi. tan (- 690°)
vii. sec 240°
viii. sec (- 855°)
ix. cosec 780°
x. cot (-1110°)
Solution:
i. sin 690° = sin (720° -30°)
Solution:
i. sin 690° = sin (720° -30°)
= sin (2 x 360° – 30°)
= – sin 30°
= −12
Ex 3.3
viii. 16 sin θ cos θ cos 2θ cos 4θ cos 8θ = sin 16θ
Solution:
L.H.S. = 16 sin θ cos θ cos 2θ cos 4θ cos 8θ
= 8(2sinθ cosθ) cos2θ cos 4θ cos 8θ
= 8sin 2θ cos 2θ cos 4θ cos 8θ
= 4(2sin 2θ cos 2θ) cos 4θ cos 8θ
= 4sin 4θ cos 4θ cos 8θ
= 2(2sin 4θ cos 4θ) cos 8θ
= 2sin 8θ cos 8θ
= sin 16θ
= R.H.S.
ix. \( = 2 cot 2x
Solution:
x. [latex]\frac{\cos x}{1+\sin x}=\frac{\cot \left(\frac{x}{2}\right)-1}{\cot \left(\frac{x}{2}\right)+1}\)
Solution:
Ex 3.4
Ex 3.5
∴ L.H.S. = – 2.cos C.cos (A – B) + 2.cos2C – 1 …[From(i)]
= – 1 – 2.cosC.[cos(A – B) – cosC]
= – 1 – 2.cos C.[cos(A – B) + cos(A + B)]
… [From (i)]
= – 1 – 2.cos C.(2.cos A.cos B)
= – 1 – 4.cos A.cos B.cos C = R.H.S.
Question 2.
sin A + sin B + sin C = 4 cos A/2 cos B/2 cos C/2
Solution:
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Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 1): Chapter 3- Trigonometry – II
Chapterwise Maharashtra Board Solutions Class 11 Arts & Science Maths (Part 1) :
- Chapter 1- Angle and its Measurement
- Chapter 2- Trigonometry – I
- Chapter 3- Trigonometry – II
- Chapter 4- Determinants and Matrices
- Chapter 5- Straight Line
- Chapter 6- Circle
- Chapter 7- Conic Sections
- Chapter 8- Measures of Dispersion
- Chapter 9- Probability