If cos 9α = sin α and 9α < 90°

(c) Given. $\cos 9 \alpha=\sin \alpha$ and $9 \alpha<90^{\circ}$ i.e., acute angle.

$\sin \left(90^{\circ}-9 \alpha\right)=\sin \alpha$  $\left[\because \cos A=\sin \left(90^{\circ}-A\right)\right]$

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

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

$\begin{array}{ll}\Rightarrow & \alpha=9^{\circ}\end{array}$

$\therefore$  $\tan 5 \alpha=\tan \left(5 \times 9^{\circ}\right)=\tan 45^{\circ}=1$ $\left[\because \tan 45^{\circ}=1\right]$