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8. Introduction to Trigonometry
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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