Advertisements
Advertisements
प्रश्न
Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
Advertisements
उत्तर
LHS = cos θ sin (90° - θ) + sin θ cos (90° - θ)
= cos θ. cos θ + sin θ. sin θ
= cos2θ + sin 2θ
= 1
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(cos A-sinA+1)/(cosA+sinA-1)=cosecA+cotA ` using the identity cosec2 A = 1 cot2 A.
Prove the following trigonometric identities:
`(1 - cos^2 A) cosec^2 A = 1`
Prove the following identities:
`1/(tan A + cot A) = cos A sin A`
Prove the following identities:
`1 - cos^2A/(1 + sinA) = sinA`
If sin A + cos A = m and sec A + cosec A = n, show that : n (m2 – 1) = 2 m
If \[\sin \theta = \frac{1}{3}\] then find the value of 9tan2 θ + 9.
If x = r sin θ cos ϕ, y = r sin θ sin ϕ and z = r cos θ, then
Prove that `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ) = 2`.
If 2sin2β − cos2β = 2, then β is ______.
The value of 2sinθ can be `a + 1/a`, where a is a positive number, and a ≠ 1.
