Question

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

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

Answer 1

Answer: (D)

$\sqrt{2}\cos \theta$

Dividing by $\sin \theta$ we get,

$\cot \theta -1=\sqrt{2}\Rightarrow \cot \theta =\sqrt{2}+1$

So, $\tan \theta =\frac{1}{\sqrt{2}+1}\Rightarrow \tan \theta =\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}or\frac{-\sqrt{2}-1}{(\sqrt{2}+1)(-\sqrt{2}-1)}\Rightarrow \tan \theta =\sqrt{2}-1or-\sqrt{2}-1\Rightarrow 1+\tan \theta =\pm \sqrt{2}\Rightarrow \frac{\cos \theta +\sin \theta }{\cos \theta }=\pm \sqrt{2}\Rightarrow \cos \theta +\sin \theta =\pm \sqrt{2}\cos \theta$

Therefore, Answer is $\sqrt{2}\cos \theta$