If $ \cos 9 \alpha=\sin \alpha $ and $ 9 \alpha<90^{\circ} $, then the value of $ \tan 5 \alpha $ is
(A) $ \frac{1}{\sqrt{3}} $
(B) $ \sqrt{3} $
(C) 1
(D) 0

Given:

\( \cos 9 \alpha=\sin \alpha \) and \( 9 \alpha<90^{\circ} \)

To do:

We have to find the value of \( \tan 5 \alpha \).

Solution:  

We know that,

$sin\ (90^{\circ}- \theta) = cos\ \theta$

$\cos 9 \alpha=\sin \alpha$

$\sin (90^{\circ}-9 \alpha)=\sin \alpha$

This implies,

$90^{\circ}-9 \alpha=\alpha$

$10 \alpha=90^{\circ}$

$\alpha=9^{\circ}$

Therefore,

$\tan 5 \alpha=\tan (5 \times 9^{\circ})$

$=\tan 45^{\circ}$

$=1$             (Since $\tan 45^{\circ}=1$)