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8. Introduction to Trigonometry
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If $\cos 9 \alpha=\sin \alpha$ and $9 \alpha<90^{\circ},$ then the value of $\tan 5 \alpha$ is
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
Which of the following pair is correct for trigonometric inter-relationship ?
| $1 .$ $\cos \theta$ | $a.$ $\frac{\cos \theta}{\sin \theta}$ |
| $2.$ $\tan \theta$ | $b.$ $\frac{1}{\operatorname{coses} \theta}$ |
| $3 .$ $\cot \theta$ | $c.$ $\frac{1}{\sec \theta}$ |
| $4.$ $\sin \theta$ | $d.$ $\frac{1}{\cot \theta}$ |
| $e.$ $\sin \theta \cdot \cos \theta$ |
easy
