Question

If $\sin \left(\theta \right)+\cos \left(\theta \right)=\sqrt{2}\cos \left(\theta \right),(\mathrm{\theta }\ne {90}^{\circ }),$, then value of $\tan \left(\theta \right)$is:

• A. $\sqrt{2}-1$
• B. $\sqrt{2}+1$
• C. $\sqrt{2}$
• D. $-\sqrt{2}$

Answer 1

The correct option is A

$\sqrt{2}-1$

Simplify the expression:

We have, $\sin \left(\theta \right)+\cos \left(\theta \right)=\sqrt{2}\cos \left(\theta \right),(\mathrm{\theta }\ne {90}^{\circ })$

Divide the expression by $\cos \left(\theta \right)$ to get the expression in terms of $\tan \left(\theta \right)$,

$\Rightarrow \frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}+\frac{\cos \left(\theta \right)}{\cos \left(\theta \right)}=\sqrt{2}\frac{\cos \left(\theta \right)}{\cos \left(\theta \right)}$

$\Rightarrow \tan \left(\theta \right)+1=\sqrt{2}[\because \frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}=\tan \left(\theta \right)]$

$\Rightarrow \tan \left(\theta \right)=\sqrt{2}-1$

Final answer: Hence, option A is correct.