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

To do:

We have to prove that \( \frac{1}{\sec A+\tan A}-\frac{1}{\cos A}=\frac{1}{\cos A}-\frac{1}{\sec A-\tan A} \).

Solution:

We know that,

$\sin^2 A+\cos^2 A=1$

$\operatorname{cosec}^2 A-\cot^2 A=1$

$\sec^2 A-\tan^2 A=1$

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

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

$\operatorname{cosec} A=\frac{1}{\sin A}$

$\sec A=\frac{1}{\cos A}$

Therefore,

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

$=\frac{1}{\frac{1+\sin A}{\cos A}}-\frac{1}{\cos A}$

$=\frac{\cos A}{1+\sin A}-\frac{1}{\cos A}$

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

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

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

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

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

$=-\tan A$

Let us consider RHS,

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

$=\frac{1}{\cos A}-\frac{1}{\frac{1-\sin A}{\cos A}}$

$=\frac{1}{\cos A}-\frac{\cos A}{1-\sin A}$

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

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

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

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

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

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

$=-\tan \mathrm{A}$

Here,

LHS $=$ RHS

Hence proved.