Advertisements
Advertisements
प्रश्न
Prove that cosec2 (90° - θ) + cot2 (90° - θ) = 1 + 2 tan2 θ.
Advertisements
उत्तर
LHS = cosec2 (90° - θ) + cot2 (90° - θ)
= sec2 θ + tan2θ
= 1 + tan2θ + tan2θ
= 1 + 2 tan2θ
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`(cos^2 theta)/sin theta - cosec theta + sin theta = 0`
Prove the following trigonometric identities.
`(1 + tan^2 A) + (1 + 1/tan^2 A) = 1/(sin^2 A - sin^4 A)`
If x = r cos A cos B, y = r cos A sin B and z = r sin A, show that : x2 + y2 + z2 = r2
`cos^2 theta /((1 tan theta))+ sin ^3 theta/((sin theta - cos theta))=(1+sin theta cos theta)`
Prove that `( sintheta - 2 sin ^3 theta ) = ( 2 cos ^3 theta - cos theta) tan theta`
Write the value of `(1 - cos^2 theta ) cosec^2 theta`.
If x = r sin θ cos ϕ, y = r sin θ sin ϕ and z = r cos θ, then
Prove the following identity :
`2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1 = 0`
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)) + sqrt((1 - sin theta)/(1 + sin theta))` = 2 sec θ
`sin θ = 1/2`, then θ = ?
