Question

Question 7
If $cos9\alpha =sin\alpha$ and $9a<{90}^{\circ }$, then the value of $tan5\alpha$ is

(A) $\frac{1}{\sqrt{3}}$
(B) $\sqrt{3}$
(C) $1$
(D) $0$

Answer 1

The answer is (C).

$cos9\alpha =sin\alpha and9a<{90}^{\circ }i.e.,acuteangle$

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

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

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

$\Rightarrow \alpha =9^{\circ }$

$\therefore tan5\alpha =tan(5\times 9^{\circ })=tan{45}^{\circ }=1\lfloor \because tan{45}^{\circ }=1\rfloor$