Prove that Sin (90° - θ) Cos (90° - θ) = Tan θ. Cos2θ.

Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos²θ.

Sum

LHS = sin (90° - θ) cos (90° - θ)
LHS = cos θ. sin θ

RHS = tan θ. cos²θ

RHS = `sin θ/cos θ` x cos²θ

RHS = cos θ. sin θ

∴ LHS = RHS
Hence proved.

shaalaa.com

Is there an error in this question or solution?

Prove the following trigonometric identities.

`(1 + cot A + tan A)(sin A - cos A) = sec A/(cosec^2 A) - (cosec A)/sec^2 A = sin A tan A - cos A cot A`

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

Prove the following identities:

`(1 + sin A)/(1 - sin A) = (cosec A + 1)/(cosec A - 1)`

Write the value of `(1 + tan^2 theta ) cos^2 theta`.

Write the value of tan1° tan 2° ........ tan 89° .

Prove the following identity :

`(1 + tan^2θ)sinθcosθ = tanθ`

Prove that `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ = 1/(sin^2 θ. cos^2 θ) - 2`.

If x sin³θ + y cos³ θ = sin θ cos θ and x sin θ = y cos θ , then show that x² + y² = 1.

tan θ cosec²θ – tan θ is equal to

Prove that `(1 + sin B)/(cos B) + (cos B)/(1 + sin B) = 2 sec B`.