Question

Prove that $(\csc A-\sin A)(\sec A-\cos A){\sec }^{2}A=\tan A$.

Answer 1

L.H.S$=(\csc A-\sin A)(\sec A-\cos A){\sec }^{2}A$

$=(\frac{1}{\sin A}-\sin A)(\frac{1}{\cos A}-\cos A){\sec }^{2}A$

$=\left(\frac{1-{\sin }^{2}A}{\sin A}\right)\left(\frac{1-{\cos }^{2}A}{\cos A}\right){\sec }^{2}A$

$=\left(\frac{{\cos }^{2}A}{\sin A}\right)\left(\frac{{\sin }^{2}A}{\cos A}\right){\sec }^{2}A$

$=\cos A\sin A\frac{1}{{\cos }^{2}A}$

$=\frac{\sin A}{\cos A}$

$=\tan A$

$=$R.H.S

Hence proved.