Question

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

Answer 1

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

Taking L.H.S., we have

$(\csc A-\sin A)(\sec A-\cos A){\sec }^{2}A$

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

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

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

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

$=\tan A$

$=$ R.H.S.

Hence proved.