Question

6. If cosθ+sin θ=√2cosθ, show that cosθ-sinθ=√2sinθ

Answer 1

Pls mark my ans as brainlest

Answer 2

Given: $\cos\theta+\sin\theta=\sqrt{2}\cos\theta$

Squaring on both the sides, we get

$(\cos\theta+\sin\theta)^2=(\sqrt{2}\cos\theta)^2\\\\\Rightarrow\ \cos^2\theta+\sin^2\theta+2\cos\theta\sin\theta=2\cos^2\theta\\\\\Rightarrow\ 1+2\cos\theta\sin\theta=2\cos^2\theta\\\\\Rightarrow\ 2\cos\theta\sin\theta=2\cos^2\theta-1$

Now, consider

$(\cos\theta-\sin\theta)^2=\cos^2\theta+\sin^2\theta-2\cos\theta\sin\theta\\\\=1-(2\cos^2\theta-1)\\\\=2-2cos^2\theta=2-2(1-\sin^2\theta)\\\\=2-2+2\sin^2\theta\\\\=2\sin^2\theta\\\\\Rightarrow\ (\cos\theta-\sin\theta)^2=2\sin^2\theta\\\\\Rightarrow\ \cos\theta-\sin\theta=\sqrt{2}\sin\theta$

Hence, proved.