Question

If $\cos \left(9\alpha \right)=\sin \left(\alpha \right)$ and $9\alpha <90°$, then the value of $\tan \left(5\alpha \right)$ is

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

Answer 1

The correct option is C

$1$

Explanation for correct option:

Find the value of the given trigonometric expression

Given,

$\cos \left(9\alpha \right)=\sin \left(\alpha \right)$ and $9\alpha <90°$

That is $9\alpha$ is a acute angle

We know that,

$\sin \left(90°-\theta \right)=\cos \left(\theta \right)$

So,

$$\begin{gathered} \cos \left(9\alpha \right)=\sin \left(\alpha \right) \\ \Rightarrow \sin \left(90°-9\alpha \right)=\sin \left(\alpha \right) \\ \Rightarrow 90°-9\alpha =\alpha \\ \Rightarrow 90°=10\alpha \\ \Rightarrow \frac{90°}{10}=\alpha \\ \Rightarrow 9°=\alpha \\ \therefore \alpha =9° \end{gathered}$$

Substituting $\alpha =9°$ in $\tan \left(5\alpha \right)$ we get

$$\begin{gathered} \tan \left(5\alpha \right)=\tan \left(5\times 9°\right) \\ =\tan \left(45°\right) \\ =1 \end{gathered}$$

Hence, $\tan \left(5\alpha \right)=1$, so, the correct option is $\mathrm{Option}\left(\mathrm{C}\right)$