Prove that ^2 - ^2 = ^2 + ^2

Question

Prove that $\sec ^2 \theta-\cos ^2 \theta=\tan ^2 \theta+\sin ^2 \theta$

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Answer

$\text { L.H.S }=\sec ^2 \theta-\cos ^2 \theta$
$ =\sec ^2 \theta-\left(1-\sin ^2 \theta\right) \quad \ldots \ldots \cdot\left[\because \sin ^2 \theta+\cos ^2 \theta=1\therefore 1-\sin ^2 \theta=\cos ^2 \theta\right]$
$=\sec ^2 \theta-1+\sin ^2 \theta \quad$
$=\tan ^2 \theta+\sin ^2 \theta \quad \ldots \ldots \cdot\left[\because 1+\tan ^2 \theta=\sec ^2 \theta\therefore \tan ^2 \theta=\sec ^2 \theta-1\right]$
$=\text { R.H.S }$
$\therefore \sec ^2 \theta-\cos ^2 \theta=\tan ^2 \theta+\sin ^2 \theta$

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