Question

If (costheta - sin theta) = √2 sin theta then prove that cos theta + sintheta ) =√ 2 costheta.


[CBSE​

Answer 1

Answer:-

Given:-

cos θ - sin θ = √2 sin θ

Squaring both sides we get,

⟹ (cos θ - sin θ)² = (√2 sin θ)²

using (a - b)² = a² + b² - 2ab we get,

⟹ cos² θ + sin² θ - 2 sin θ cos θ = 2 sin² θ

⟹ cos² θ = 2 sin² θ - sin² θ + 2 sin θ cos θ

Now, Adding cos² θ both sides we get,

⟹ cos² θ + cos² θ = sin² θ + cos² θ + 2 sin θ cos θ

using a² + b² + 2ab = (a + b)² in RHS we get,

⟹ 2 cos² θ = (sin θ + cos θ)²

⟹ √(2 cos² θ) = √(sin θ + cos θ)²

Square and root get cancelled both sides.

⟹ √2 cos θ = sin θ + cos θ

Hence, Proved.

Answer 2

Given :-

If  (cos theta - sin theta) = √2 sin theta

To Find :-

Prove that

(cos theta + sin theta ) = √2 cos theta.

Solution :-

At first we need to square both the sides

$\sf \bigg(cos \theta - sin \theta\bigg)^2 = \sf\bigg(\sqrt{2} \; sin \theta\bigg)^2$

$\sf cos^2 \theta + sin^2 \theta - 2 \; sin \theta \times cos\theta =( \sqrt{2} \; sin\theta)^2$

$\sf cos^2 \theta + sin^2 \theta - 2 \; sin \theta \times cos\theta = 2 \; sin^2\theta$

$\sf cos^2 \theta + cos^2 \theta = sin^2 \theta + cos^2 \theta + 2\; sin \theta\; cos \theta$

$\sf 2(cos^2\theta) = sin^2\theta + cos^2\theta + 2\sin\theta cos\theta$

$\sf 2(cos \theta)^2=sin\theta^2 +cos\theta^2$

$\sf \sqrt{2(cos \theta)^2}=\sqrt{sin\theta^2 +cos\theta^2}$

$\sf\sqrt{2}cos\theta = sin\theta+cos\theta$