Chapter 3 Trigonometry – II Ex 3.2 Solutions

Maharashtra State Board 11th Maths Solutions Chapter 3 Trigonometry – II Ex 3.2

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°
= –

ii. L.H.S.
= cos ( + x) cos (2π + x). [cot ( – x) + (2π + x)]
= (sin x)(cos x) (tan x + cot x)
= sin x cos x ( )
= sin x cos x
= sin x cos x
= 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°)
= – sin 60° = –

tan (-690°) = – tan 690°
= – tan (360° +330°)
= -tan (330°)
=- tan (360° – 30°)
=-(-tan 30°)
= tan 30°0 =

L.H.S. = sec 840° cot (-945°) + sin 600° tan (-690°)
= (-2)(-1) +
= 2 –
= 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.