# 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),

$(1+\cot A+\tan A)(\sin A-\cos A)=\frac{\sec A}{\operatorname{cosec}^{2} A}-\frac{\operatorname{cosec} A}{\sec ^{2} A}=\sin A \tan A-\cot A \cos A$.

Hence proved.

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Updated on: 10-Oct-2022

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