Prove that Sec2 (90° - θ) + Tan2 (90° - θ) = 1 + 2 Cot2 θ.

Prove that sec² (90° - θ) + tan² (90° - θ) = 1 + 2 cot² θ.

Sum

LHS = sec² (90° - θ) + tan² (90° - θ)

= cosec²θ + cot²θ

= 1 + cot²θ + cot²θ

= 1 + 2cot²θ

= RHS
Hence proved.

shaalaa.com

Is there an error in this question or solution?

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

`(sintheta - 2sin^3theta)/(2costheta - costheta) =tan theta`

Prove the following trigonometric identities.

`((1 + sin theta - cos theta)/(1 + sin theta + cos theta))^2 = (1 - cos theta)/(1 + cos theta)`

Prove the following trigonometric identities.

`(tan A + tan B)/(cot A + cot B) = tan A tan B`

Prove the following identities:

`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`

Prove that:

(1 + tan A . tan B)² + (tan A – tan B)² = sec² A sec² B

If x= a sec `theta + b tan theta and y = a tan theta + b sec theta ,"prove that" (x^2 - y^2 )=(a^2 -b^2)`

If `sec theta = x ,"write the value of tan" theta`.

Prove that `sqrt((1 - sin θ)/(1 + sin θ)) = sec θ - tan θ`.

Prove that: `(sec θ - tan θ)/(sec θ + tan θ ) = 1 - 2 sec θ.tan θ + 2 tan^2θ`

a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p² – q² is equal to