Advertisements
Advertisements
प्रश्न
Evaluate the following: sin(35° + θ) - cos(55° - θ) - tan(42° + θ) + cot(48° - θ)
Advertisements
उत्तर
sin(35° + θ) - cos(55° - θ) - tan(42° + θ) + cot(48° - θ)
= sin[90° - (55° - θ)] - cos(55° - θ) - tan[90° - (48° - θ)] + cot(48° - θ)
= cos(55° - θ) - cos(55° - θ) - cot(48° - θ) + cot(48° - θ)
= 0.
APPEARS IN
संबंधित प्रश्न
If 2 sin x° − 1 = 0 and x° is an acute angle; find:
- sin x°
- x°
- cos x° and tan x°.
If sin 3A = 1 and 0 < A < 90°, find `tan^2A - (1)/(cos^2 "A")`
Find the magnitude of angle A, if 2 cos2 A - 3 cos A + 1 = 0
Solve for 'θ': `sin θ/(3)` = 1
If `sqrt(3)` sec 2θ = 2 and θ< 90°, find the value of
cos2 (30° + θ) + sin2 (45° - θ)
In right-angled triangle ABC; ∠B = 90°. Find the magnitude of angle A, if:
a. AB is `sqrt(3)` times of BC.
B. BC is `sqrt(3)` times of BC.
Evaluate the following: `(sec32° cot26°)/(tan64° "cosec"58°)`
Express each of the following in terms of trigonometric ratios of angles between 0° and 45°: cos84° + cosec69° - cot68°
If sin(θ - 15°) = cos(θ - 25°), find the value of θ if (θ-15°) and (θ - 25°) are acute angles.
Prove the following: `(tan(90° - θ)cotθ)/("cosec"^2 θ)` = cos2θ
