Question

Prove that tan²θ + cos²θ – 1 = tan²θ. sin²θ

Answer 1

L.H.S. = tan²θ + cos²θ – 1

= tan²θ – (1 – cos²θ)

= `(sin^2θ)/(cos^2θ) - sin^2θ`

= `(sin^2θ (1 - cos^2θ))/(cos^2θ)`

= `(sin^2θ)/(cos^2θ) xx sin^2θ`

= tan²θ. sin²θ = R.H.S.