Solveeit Logo
Login

Question

Question: If \(\cos\theta - \sin\theta = \sqrt{2}\sin\theta,\) then \(\cos\theta + \sin\theta\)is equal to...

If cos⁡θ−sin⁡θ=2sin⁡θ,\cos\theta - \sin\theta = \sqrt{2}\sin\theta, then cos⁡θ+sin⁡θ\cos\theta + \sin\thetais equal to

A

2cos⁡θ\sqrt{2}\cos\theta

B

2sin⁡θ\sqrt{2}\sin\theta

C

2cos⁡θ2\cos\theta

D

−2cos⁡θ- \sqrt{2}\cos\theta

Answer

2cos⁡θ\sqrt{2}\cos\theta

Explanation

Solution

We have cos⁡θ−sin⁡θ=2sin⁡θ\cos\theta - \sin\theta = \sqrt{2}\sin\theta

⇒cos⁡θ=(2+1)sin⁡θ⇒(2−1)cos⁡θ=sin⁡θ\Rightarrow \cos\theta = (\sqrt{2} + 1)\sin\theta \Rightarrow (\sqrt{2} - 1)\cos\theta = \sin\theta

⇒2cos⁡θ−cos⁡θ=sin⁡θ⇒sin⁡θ+cos⁡θ=2cos⁡θ.\Rightarrow \sqrt{2}\cos\theta - \cos\theta = \sin\theta \Rightarrow \sin\theta + \cos\theta = \sqrt{2}\cos\theta.