Question

prove that tan squared theta minus sin square theta is equal to tan square theta sin square theta ​

Answer 1

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Answer 2

$\tan^2 \theta-\sin^2 \theta=\tan^2 \theta \sin^2 \theta$, proved

Step-by-step explanation:

To prove that $\tan^2 \theta-\sin^2 \theta=\tan^2 \theta \sin^2 \theta$.

L.H.S. = $\tan^2 \theta-\sin^2 \theta$

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

Using the trigonometric identity,

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

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

$=\dfrac{\sin^2 \theta(1-\cos^2 \theta)}{\cos^2 \theta}$

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

Using the trigonometric identity,

$\sin^2 A=1-\cos^2 A$

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

$=\tan^2 \theta \sin^2 \theta​$

=  R.H.S., proved.

Thus, $\tan^2 \theta-\sin^2 \theta=\tan^2 \theta \sin^2 \theta$, proved