Question
Prove that: $(cosecA - sinA) (secA - cosA) sec^2A = tanA$
✓
Answer
$ \text{LHS} \ (cosecA - sinA) (secA - cosA) sec^2A$
$=\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right) \sec ^2 A$
$=\left(\frac{1-\sin ^2 A}{\sin A}\right)\left(\frac{1-\cos ^2 A}{\cos A}\right) \sec ^2 A$
$=\left(\frac{\cos ^2 A}{\sin A}\right)\left(\frac{\sin ^2 A}{\cos A}\right) \sec ^2 A$
$=\frac{\sin A}{\cos A}=\tan A= \text{RHS}$
$=\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right) \sec ^2 A$
$=\left(\frac{1-\sin ^2 A}{\sin A}\right)\left(\frac{1-\cos ^2 A}{\cos A}\right) \sec ^2 A$
$=\left(\frac{\cos ^2 A}{\sin A}\right)\left(\frac{\sin ^2 A}{\cos A}\right) \sec ^2 A$
$=\frac{\sin A}{\cos A}=\tan A= \text{RHS}$
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