Chapter 3 Trigonometry – II Ex 3.1 Solutions

Maharashtra State Board 11th Maths Solutions Chapter 3 Trigonometry – II Ex 3.1

ii. cos 75° = cos (45° + 30°)
= cos 45° cos 30° – sin 45° sin 30°

iii. tan 105° = tan (60° +45°)

iv. cot 225°

ii.
L.H.S =

R.H.S.
[Note : The question has been modified.]

iii.
Solution:

iv. sin [(n+1)A]. sin [(n+2)A] + cos [(n+1)A]. cos [(n+2)A] = cos A
Solution:
L.H.S. = sin [(n + 1)A]. sin [(n + 2)A] + cos [(n + 1)A]. cos [(n + 2)A]
= cos [(n + 2)A]. cos [(n + 1)A] + sin [(n + 2)A]. sin [(n + 1)A]
Let(n+2)Aaand(n+l)Ab …(i)
∴ L.H.S. = cos a. cos b + sin a. sin b
= cos (a — b)
= cos [(n + 2)A — (n + I )A]
…[From (i)]
cos[(n+2 – n – 1)A]
= cos A
= R.H.S.

v.
Solution:

vi.
Solution:

vii. cos (x + y). cos (x – y) = cos²y – sin²x
Solution:
L.H.S. = cos(x + y). cos(x – y)
= (cos x cos y – sin x sin y). (cos x cos y + sin x sin y)
= cos²x cos²y – sin²x sin²y
…[∵ (a – b) (a + b) = a²– b²]
= (1 – sin²x) cos²y – sin²x (1 – cos²y)
…[∵ sin²e + cos²0 = 1]
= cos²y – cos²y sin²x – sin²x + sin²x cos²y
= cos²y – sin²x
=R.H.S.

viii.
Solution:

ix. tan 8θ – tan 5θ – tan 3θ = tan 8θ tan 5θ tan 3θ
Solution:
Since, 8θ = 5θ + 3θ
∴ tan 8θ = tan (5θ + 3θ)
∴ tan 8θ =
∴ tan 8θ (1 – tan 5θ.tan 3θ) = tan 5θ + tan 3θ
∴ tan 8θ – tan8θ.tan5θ.tan3θ = tan5θ + tan 3θ
∴ tan 8θ – tan 5θ – tan 3θ = tan 8θ.tan 5θ.tan 3θ

x. tan 50° = tan 40° + 2tan 10°
Solution:
Since, 50° = 10° +40°
∴ tan 50° = tan (10° + 40°)
∴
∴ tan 50° (1 – tan 10° tan 40°) = tan 10° + tan 40°
∴ tan 50° – tan 10° tan 40° tan 50° = tan 10° + tan 40°
∴ tan 50° – tan 10° tan 40° tan (90° – 40°) = tan 10° + tan 40°
∴ tan 50° – tan 10° tan 40° cot 40°
= tan 10° + tan 40° …[∵ tan (90° – θ) = cot θ]
∴ tan 50° – tan 10° tan 40°. = tan 10° + tan 40°
∴ tan 50° – tan 10°. 1 = tan 10° + tan 40°
∴ tan 50° = tan 40° + 2 tan 10°

xi. = tan 72°
Solution:

Dividing numerator and cos 27°, we get denominator by cos 27°, we get

= tan (45° + 27°)
= tan 72° = R.H.S

xii.
Solution:
Since 45° = 10° + 35°,
tan 45° = tan (10° +35°)
∴
∴ 1 – tan 10° tan 35o = tan 10° + tan 35°
∴ tan 10° + tan 35° + tan 10° tan 35° = 1

xiii. tan 10° + tan 35° + tan 10°. tan 35° = 1
Solution:

xiv.
Solution:
Dividing numerator and cos 15°, we get

= tan (45° + 15°)
= tan 30° = = R.H.S

i. sin (A + B) = sin A cos B+cos A sin B

ii. cos (A -B) = cos A cos B + sin A sin B

iii.