Question

If $\cos 9\alpha =\sin \alpha$, and $9\alpha <{90}^{\circ }$, then $\tan 5\alpha =.....$

• A. $\frac{1}{\sqrt{3}}$
• B. $\sqrt{3}$
• C. $1$
• D. $0$

Answer 1

The correct option is C $1$

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

So, $\cos 9\alpha =\cos (90-\alpha )\therefore 9\alpha =90-\alpha \Rightarrow 9\alpha +\alpha ={90}^{\circ }\Rightarrow 10\alpha ={90}^o\Rightarrow \alpha =9^o$

Now, $\tan 5\alpha =\tan (5\ast 9^o)=\tan {45}^o=1$

Therefore, Answer is $1$