Question

find the value of sin theta/ root 1- sin square theta

Answer 1

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Answer 2

$\dfrac{\sin \theta}{\sqrt{1-\sin^2 \theta}}=\tan \theta$

Step-by-step explanation:

We have,

$\dfrac{\sin \theta}{\sqrt{1-\sin^2 \theta}}$

To find, $\dfrac{\sin \theta}{\sqrt{1-\sin^2 \theta}}=?$

∴ $\dfrac{\sin \theta}{\sqrt{1-\sin^2 \theta}}$

$=\dfrac{\sin \theta}{\sqrt{\cos^2 \theta}}$

Using trigonometric identity,

$\sin^2 \theta+\cos^2 \theta=1$

$=\dfrac{\sin \theta}{\cos \theta}$

$=\tan \theta$

Using trigonometric identity,

$\tan \theta=\dfrac{\sin \theta}{\cos \theta}$

Hence, $\dfrac{\sin \theta}{\sqrt{1-\sin^2 \theta}}=\tan \theta$