Evaluate the following: sin31° - cos59°

बेरीज

sin31° - cos59°
= sin(90° - 59°) - cos59°
= cos59° - cos59°
= 0.

shaalaa.com

Trigonometric Equation Problem and Solution

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

पाठ 22: Trigonometrical Ratios of Standard Angles - Exercise 27.3

पाठ 22 Trigonometrical Ratios of Standard Angles
Exercise 27.3 | Q 2.1

If 2 cos 2A = `sqrt3` and A is acute,
find:
(i) A
(ii) sin 3A
(iii) sin² (75° - A) + cos² (45° +A)

Solve for x : cos² 30° + cos² x = 1

Solve for 'θ': cot²(θ - 5)° = 3

If θ = 30°, verify that: sin 3θ = 4sinθ . sin(60° - θ) sin(60° + θ)

If `sqrt(3)` sec 2θ = 2 and θ< 90°, find the value of

cos² (30° + θ) + sin² (45° - θ)

In the given figure, PQ = 6 cm, RQ = x cm and RP = 10 cm, find
a. cosθ
b. sin²θ- cos²θ
c. Use tanθ to find the value of RQ

Find the value of: `sqrt((1 - sin^2 60°)/(1 + sin^2 60°)` If 3 tan²θ - 1 = 0, find the value
a. cosθ
b. sinθ

Find x and y, in each of the following figure:

Evaluate the following: `(sin0° sin35° sin55° sin75°)/(cos22° cos64° cos58° cos90°)`

Prove the following: `(tan(90° - θ)cotθ)/("cosec"^2 θ)` = cos²θ