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Prove the following identities:$ \frac{\sin A}{\sec A+\tan A-1}+\frac{\cos A}{\operatorname{cosec} A+\cot A-1}=1 $
To do:
We have to prove that \( \frac{\sin A}{\sec A+\tan A-1}+\frac{\cos A}{\operatorname{cosec} A+\cot A-1}=1 \).
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,
$\frac{\sin A}{\sec A+\tan A-1}+\frac{\cos A}{\operatorname{cosec} A+\cot A-1}=\frac{\sin A}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}-1}+\frac{\cos A}{\frac{1}{\sin A}+\frac{\cos A}{\sin A}-1}$
$=\frac{\sin A}{\frac{1+\sin A-\cos A}{\cos A}}+\frac{\cos A}{\frac{1+\cos A-\sin A}{\sin A}}$
$=\frac{\sin A \cos A}{1+\sin A-\cos A}+\frac{\sin A \cos A}{1-\sin A+\cos A}$
$=\sin A \cos A \left(\frac{1}{1+\sin A-\cos A}+\frac{1}{1-\sin A+\cos A}\right)$
$=\sin A \cos A \left(\frac{1-\sin A+\cos A+1+\sin A-\cos A}{(1+\sin A-\cos A)(1-\sin A+\cos A)}\right)$
$=\sin A \cos A \left(\frac{2}{(1+\sin A-\cos A)(1-\sin A+\cos A)}\right)$
$= \frac{2\sin A \cos A}{(1)^{2}-(\sin A-\cos A)^{2}}$
$=\frac{2 \sin A \cos A}{1-\left(\sin ^{2} A+\cos ^{2} A-2 \sin A \cos A\right)}$
$=\frac{2 \sin A \cos A}{1-(1-2 \sin A \cos A)}$
$=\frac{2 \sin A \cos A}{1-1+2 \sin A \cos A}$
$=\frac{2 \sin A \cos A}{2 \sin A \cos A}$
$=1$
Hence proved.
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