Question
Prove the following trigonometric identities.
$(\sec\theta+\cos\theta)(\sec\theta-\cos\theta)=\tan^2\theta+\sin^2\theta$
$(\sec\theta+\cos\theta)(\sec\theta-\cos\theta)=\tan^2\theta+\sin^2\theta$
✓
Answer
We have to prove $(\sec\theta+\cos\theta)(\sec\theta-\cos\theta)=\tan^2\theta+\sin^2\theta$
We know that, $\sin^2\theta+\cos^2\theta=1$
$\text{L.H.S}=(\sec\theta+\cos\theta)(\sec\theta-\cos\theta)$
$\sec^2\theta-\tan^2\theta=0$
$(\sec\theta+\cos\theta)(\sec\theta-\cos\theta)=\sec^2\theta-\cos^2\theta$
$=(1+\tan^2\theta)-(1-\sin^2\theta)$
$1+\tan^2\theta-1+\sin^2\theta$
$=\tan^2\theta+\sin^2\theta=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
We know that, $\sin^2\theta+\cos^2\theta=1$
$\text{L.H.S}=(\sec\theta+\cos\theta)(\sec\theta-\cos\theta)$
$\sec^2\theta-\tan^2\theta=0$
$(\sec\theta+\cos\theta)(\sec\theta-\cos\theta)=\sec^2\theta-\cos^2\theta$
$=(1+\tan^2\theta)-(1-\sin^2\theta)$
$1+\tan^2\theta-1+\sin^2\theta$
$=\tan^2\theta+\sin^2\theta=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
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