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Given:

To Prove: a sin 2θ + b cos 2θ = b

Given:

We know that,

By Pythagoras Theorem,

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (a)² + (b)² = (H)²

⇒ a² + b² = (H)²

So,

Taking LHS,

= a sin 2θ + b cos 2θ

We know that,

sin 2θ = 2 sin θ cos θ

and cos 2θ = 1 – 2 sin²θ

= a(2 sin θ cos θ) + b(1 – 2 sin²θ)

Putting the values of sinθ and cosθ, we get

= b

= RHS

∴ LHS = RHS

Hence Proved