If cos 9α = sin α and 9α < 90°

Question:

If cos 9α = sin α and 9α < 90° ,then the value of tan 5α is

(a) $\frac{1}{\sqrt{3}}$

(b) $\frac{\sqrt{3}}{1}$

(C) 7

(d) 0

Solution:

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

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

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

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

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

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

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