Question
Prove that $\cot ^2 \theta-\tan ^2 \theta=\operatorname{cosec}^2 \theta-\sec ^2 \theta$
✓
Answer
$\text { L.H.S }=\cot ^2 \theta-\tan ^2 \theta$
$=\left(\operatorname{cosec}^2 \theta-1\right)-\left(\sec ^2 \theta-1\right) \quad \cdots \cdot\left[\because \tan ^2 \theta=\sec ^2 \theta-1\cot ^2 \theta=\operatorname{cosec}^2 \theta-1\right]$
$=\operatorname{cosec}^2 \theta-1-\sec ^2 \theta+1$
$=\operatorname{cosec}^2 \theta-\sec ^2 \theta$
$=\text { R.H.S }$
$\therefore \cot ^2 \theta-\tan ^2 \theta=\operatorname{cosec}^2 \theta-\sec ^2 \theta$
$=\left(\operatorname{cosec}^2 \theta-1\right)-\left(\sec ^2 \theta-1\right) \quad \cdots \cdot\left[\because \tan ^2 \theta=\sec ^2 \theta-1\cot ^2 \theta=\operatorname{cosec}^2 \theta-1\right]$
$=\operatorname{cosec}^2 \theta-1-\sec ^2 \theta+1$
$=\operatorname{cosec}^2 \theta-\sec ^2 \theta$
$=\text { R.H.S }$
$\therefore \cot ^2 \theta-\tan ^2 \theta=\operatorname{cosec}^2 \theta-\sec ^2 \theta$
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