Question
Prove that $(cosec A - \sin A)( \sec A - \cos A) \sec^2 A = tan A$.

Answer

LHS = $(cosec A - \sin A)(\sec A - \cos A). \sec^2A$
$=\left(\frac{1}{\sin A}-\sin A\right) \cdot\left(\frac{1}{\cos A}-\cos A\right) \cdot \frac{1}{\cos ^2 A}$
$=\frac{1-\sin ^2 A}{\sin A} \cdot \frac{1-\cos ^2 A}{\cos A} \times \frac{1}{\cos ^2 A}$
$=\frac{\cos ^2 A}{\sin A} \times \frac{\sin ^2 A}{\cos A} \times \frac{1}{\cos ^2 A} \ldots\left[\because\left(1-\sin ^2 A\right)=\cos ^2 A, 1-\cos ^2 A=\sin ^2 A\right]$
$ =\frac{\sin A}{\cos A}=\tan A$
$= RHS $
Hence proved.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free