Question

If $\cos \theta -\sin \theta =\sqrt{2}\sin \theta$, then $\cos \theta +\sin \theta$ is equal to

• A. $\sqrt{2}\cos \theta$
• B. $\sqrt{2}\sin \theta$
• C. $2\cos \theta$
• D. $-\sqrt{2}\cos \theta$

Answer 1

The correct option is A

$\sqrt{2}\cos \theta$

Step 1. Find the value of $\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\theta }$:

Given, $\cos \theta -\sin \theta =\sqrt{2}\sin \theta$

square both sides, we get

${\cos }^2\theta +{\sin }^2\theta –2\sin \theta \cos \theta =2{\sin }^2\theta$

$\Rightarrow$ $1–2\sin \theta \cos \theta =2{\sin }^2\theta$

$\Rightarrow$ $2\sin \theta \cos \theta =1–2{\sin }^2\theta$ ….(i)

Now, consider $\cos \theta +\sin \theta$

Step 2. By Squaring this expression, we get

$\begin{array}{rcl}{(cos\theta +sin\theta )}^2& =& co{s}^2\theta +si{n}^2\theta +2sin\theta cos\theta \\ & =& 1+1–2si{n}^2\theta \\ & =& 2–2si{n}^2\theta \\ & =& 2(1–si{n}^2\theta )\\ & =& 2co{s}^2\theta \end{array}$

$\mathbf{\therefore }\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\sqrt{\mathbf{2}}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }$

Hence, Option ‘A’ is Correct.