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Prove the following identities:$ (1+\cot A+\tan A)(\sin A-\cos A)=\frac{\sec A}{\operatorname{cosec}^{2} A}-\frac{\operatorname{cosec} A}{\sec ^{2} A}=\sin A \tan A-\cot A \cos A $
To do:
We have to prove that \( (1+\cot A+\tan A)(\sin A-\cos A)=\frac{\sec A}{\operatorname{cosec}^{2} A}-\frac{\operatorname{cosec} A}{\sec ^{2} A}=\sin A \tan A-\cot A \cos A \).
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,
$(1+\cot A+\tan A)(\sin A-\cos A)=\left(1+\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}\right)(\sin A-\cos A)$
$=(\frac{\sin A \cos A+\cos ^{2} A+\sin ^{2} A}{\sin A \cos A})(\sin A-\cos A)$
$=\frac{(\sin A-\cos A)\left(\sin ^{2} A+\sin A \cos A+\cos ^{2} A\right)}{\sin A \cos A}$
$=\frac{\sin ^{3} A-\cos ^{3} A}{\sin A \cos A}$.......(i)
$\frac{\sec A}{\operatorname{cosec}^{2} A}-\frac{\operatorname{cosec} A}{\sec ^{2} A}=\frac{\sin ^{2} A}{\cos A}-\frac{\cos ^{2} A}{\sin A}$
$=\frac{\sin ^{3} \mathrm{~A}-\cos ^{3} \mathrm{~A}}{\sin \mathrm{A} \cos \mathrm{A}}$............(ii)
$\sin A \tan A-\cot A \cos A=\sin A (\frac{\sin A}{\cos A})-(\frac{\cos A}{\sin A}) \cos A$
$=\frac{\sin ^{2} A}{\cos A}-\frac{\cos ^{2} A}{\sin A}$
$=\frac{\sin ^{3} A-\cos ^{3} A}{\sin A \cos A}$.......(iii)
From (i), (ii) and (iii),
\( (1+\cot A+\tan A)(\sin A-\cos A)=\frac{\sec A}{\operatorname{cosec}^{2} A}-\frac{\operatorname{cosec} A}{\sec ^{2} A}=\sin A \tan A-\cot A \cos A \).
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