Question

Show that:
$\frac{1}{\sec A+\tan A}-\frac{1}{\cos A}=\frac{1}{\cos A}-\frac{1}{\sec A-\tan A}$

Answer 1

To prove:- $\frac{1}{\sec A+\tan A}-\frac{1}{\cos A}=\frac{1}{\cos A}-\frac{1}{\sec A-\tan A}$

Proof:-

$\frac{1}{\sec A+\tan A}-\frac{1}{\cos A}=\frac{1}{\cos A}-\frac{1}{\sec A-\tan A}$

$\Rightarrow \frac{1}{\sec A+\tan A}+\frac{1}{\sec A-\tan A}=\frac{1}{\cos A}+\frac{1}{\cos A}$

$\Rightarrow \frac{\sec A-\tan A+\sec A+\tan A}{(\sec A-\tan A)(\sec A+\tan A)}=\frac{1+1}{\cos A}$

$\Rightarrow \frac{2\sec A}{{\sec }^{2}A-{\tan }^{2}A}=\frac{2}{\cos A}$

$\Rightarrow \frac{2\sec A}{1}=2\sec A(\because 1+{\tan }^{2}A={\sec }^{2}A)$

$\Rightarrow 2\sec A=2\sec A$

$\Rightarrow$ L.H.S. $=$ R.H.S.

Hence proved.