Question

prove that sin 90_ theta .cos 90_ theta=tan theta / 1+ tan square theta

Answer 1

Answer :

Now, R.H.S.

= tanθ/(1 + tan²θ)

= tanθ/sec²θ, since sec²θ - tan²θ = 1

= tanθ cos²θ, since secθ = 1/cosθ

= sinθ/cosθ × cos²θ, since tanθ = sinθ/cosθ

= cosθ sinθ

= sin(90° - θ) cos(90° - θ)

= L.H.S.

⇒ sin(90° - θ) cos(90° - θ) = tanθ/(1 + tan²θ) [Proved]

#MarkAsBrainliest