8. Introduction to Trigonometry
medium

यदि $\cos 9 \alpha=\sin \alpha$ है और $9 \alpha<90^{\circ}$ है, तो $\tan 5 \alpha$ का मान है

A

$\frac{1}{\sqrt{3}}$

B

$\sqrt{3}$

C

$1$

D

$0$

Solution

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

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

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

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

$\alpha=9^{\circ}$

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

Standard 10
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.