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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.
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