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Prove the following identities:$ \frac{\cot ^{2} A(\sec A-1)}{1+\sin A}=\sec ^{2} A\left(\frac{1-\sin A}{1+\sec A}\right) $
To do:
We have to prove that \( \frac{\cot ^{2} A(\sec A-1)}{1+\sin A}=\sec ^{2} A\left(\frac{1-\sin A}{1+\sec A}\right) \).
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,
Let us consider LHS,
$\frac{\cot ^{2} A(\sec A-1)}{1+\sin A}=\frac{\cos ^{2} A\left(\frac{1}{\cos A}-1\right)}{\sin ^{2} A(1+\sin A)}$
$=\frac{\cos ^{2} A(1-\cos A)}{\cos A \sin ^{2} A (1+\sin A)}$
$=\frac{\cos A(1-\cos A)}{\left(1-\cos ^{2} A\right)(1+\sin A)}$
$=\frac{\cos A(1-\cos A)}{(1+\cos A)(1-\cos A)(1+\sin A)}$
$=\frac{\cos A}{(1+\sin A)(1+\cos A)}$
Let us consider RHS,
$\sec ^{2} A\left(\frac{1-\sin A}{1+\sec A}\right)=\frac{1}{\cos ^{2} \mathrm{~A}}\left(\frac{1-\sin \mathrm{A}}{1+\frac{1}{\cos \mathrm{A}}}\right)$
$=\frac{1}{\cos ^{2} \mathrm{~A}}(\frac{1-\sin \mathrm{A}}{\frac{\cos \mathrm{A}+1}{\cos \mathrm{A}}})$
$=\frac{\cos \mathrm{A}(1-\sin \mathrm{A})}{\cos ^{2} \mathrm{~A}(1+\cos \mathrm{A})}$
$=\frac{\cos \mathrm{A}(1-\sin \mathrm{A})}{\left(1-\sin ^{2} \mathrm{~A}\right)(1+\cos \mathrm{A})}$
$=\frac{\cos \mathrm{A}(1-\sin \mathrm{A})}{(1+\sin \mathrm{A})(1-\sin \mathrm{A})(1+\cos \mathrm{A})}$
$=\frac{\cos \mathrm{A}}{(1+\sin \mathrm{A})(1+\cos \mathrm{A})}$
Here,
LHS $=$ RHS
Hence proved.
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