Question
Prove that $\tan^2\Phi + \cot^2\Phi + 2 = sec^2\Phi .cosec^2\Phi .$
✓
Answer
$L.H.S. = \tan^2\Phi + \cot^2\Phi + 2$
$= \tan^2\Phi + 1 + \cot^2\Phi + 1$
$= sec^2\Phi + cosec^2\Phi$
$=\frac{1}{\cos ^2 \Phi}+\frac{1}{\sin ^2 \Phi}$
$=\frac{\sin ^2 \Phi+\cos ^2 \Phi}{\sin ^2 \Phi \cdot \cos ^2 \Phi}$
$=\frac{1}{\sin ^2 \Phi \cdot \cos ^2 \Phi}$
$= cosec^2\Phi . sec^2\Phi$
$= R.H.S.$
Hence proved.
$= \tan^2\Phi + 1 + \cot^2\Phi + 1$
$= sec^2\Phi + cosec^2\Phi$
$=\frac{1}{\cos ^2 \Phi}+\frac{1}{\sin ^2 \Phi}$
$=\frac{\sin ^2 \Phi+\cos ^2 \Phi}{\sin ^2 \Phi \cdot \cos ^2 \Phi}$
$=\frac{1}{\sin ^2 \Phi \cdot \cos ^2 \Phi}$
$= cosec^2\Phi . sec^2\Phi$
$= R.H.S.$
Hence proved.
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