**Chapter 3 Trigonometry – II Ex 3.2**

## Chapter 3 Trigonometry – II 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°

= -1/2

ii. sin 495° = sin (360° + 135°)

= sin (135°)

= sin (90° + 45°)

= cos 45°

iii. cos 315° = cos (270° + 45°)

sin 45°

iv. cos 600° = cos (360° + 240°)

= cos 240°

= cos (180° + 60°)

= – cos 60°

v. tan 225° = tan (180° + 45°)

= tan 45°

= 1.

vi. tan (- 690°) = – tan 690°

= – tan (720° – 30°)

= – tan (2 x 360° – 30°)

= – (- tan 30°)

= tan 30°

vii. sec 240° = sec (180° + 60°)

= – sec 60°

= – 2

viii. sec (-855°) = sec (855°)

= sec (720°+135°)

= sec (2 x360°+ 135°) = sec 135°

= sec (90° + 45°)

= – cosec 45°

=

ix. cosec 780° = cosec (720° + 60°)

= cosec (2 x 360° + 60°)

= cosec 60°

x. cot (-1110°) =-cot (1110°)

= -cot (1080°+ 30°)

= – cot (3 x 360° + 30° )

= – cot 30°

=

**Question 2.vi. cos θ + sin (270° + θ) – sin (270° – θ) + cos (180° + θ) = 0Solution:**

i.

ii. L.H.S.

= 1 = R.H.S

iii. sec 840° = sec (720° + 120°)

= sec (2 x 360° + 120°)

= sec (120°)

= sec (90° + 30°)

= – cosec 30°

= -2

cot(-945°) = -cot 945°

= -cot (720° + 225°)

= -cot (2 x 360° +225°)

= -cot (225°)

= -cot (180° + 459)

= -cot 45°

= -1

sin 600° = sin (360° + 240°)

= sin (240°)

= sin (180° +60°)

tan (-690°) = – tan 690°

= – tan (360° +330°)

= -tan (330°)

=- tan (360° – 30°)

=-(-tan 30°)

L.H.S. = sec 840° cot (-945°) + sin 600° tan (-690°)

= R. H. S.

iv.

= 1

= R.H.S

v.

vi. L.H.S. = cos θ + sin (270° + θ) – sin (270° – θ) + cos (180° + θ)

= cos θ + (- cos θ)-(- cos θ) – cos θ

= cos θ – cos θ + cos θ – cos θ

= 0

= R.H.S.