Question

If $\sin \theta -\cos \theta =\sqrt{2}\sin ({90}^o-\theta )$, then $\tan \theta =$

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

Answer 1

The correct option is D $\sqrt{2}+1$

Given, $\sin \theta -\cos \theta =\sqrt{2}\sin (90-\theta )$
$\sin \theta -\cos \theta =\sqrt{2}\cos \theta$
$\sin \theta =\cos \theta (1+\sqrt{2})$
$\tan \theta =(\sqrt{2}-1)$