Prove the following trigonometric identities.

Solution:

We have prove that

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

We know that, $\sin ^{2} A+\cos ^{2} A=1$

So,

$(1+\cot A+\tan A)(\sin A-\cos A)$

$=\left(1+\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}\right)(\sin A-\cos A)$

$=\left(\frac{\sin A \cos A+\cos ^{2} A+\sin ^{2} A}{\sin A \cos A}\right)(\sin A-\cos A)$

$=\left(\frac{\sin A \cos A+1}{\sin A \cos A}\right)(\sin A-\cos A)$

$=\frac{(\sin A-\cos A)(\sin A \cos A+1)}{\sin A \cos A}$

$=\frac{\sin ^{2} A \cos A+\sin A-\cos ^{2} A \sin A-\cos A}{\sin A \cos A}$

$=\frac{\left(\sin ^{2} A \cos A-\cos A\right)+\left(\sin A-\cos ^{2} A \sin A\right)}{\sin A \cos A}$

$=\frac{\cos A\left(\sin ^{2} A-1\right)+\sin A\left(1-\cos ^{2} A\right)}{\sin A \cos A}$

$=\frac{\cos A\left(-\cos ^{2} A\right)+\sin A\left(\sin ^{2} A\right)}{\sin A \cos A}$

$=\frac{-\cos ^{3} A+\sin ^{3} A}{\sin A \cos A}$

$=\frac{\sin ^{3} A-\cos ^{3} A}{\sin A \cos A}$

$=\frac{\sin ^{3} A}{\sin A \cos A}-\frac{\cos ^{3} A}{\sin A \cos A}$

$=\frac{\sin ^{2} A}{\cos A}-\frac{\cos ^{2} A}{\sin A}$

$=\frac{\sin A}{\cos A} \sin A-\frac{\cos A}{\sin A} \cos A$

$=\tan A \sin A-\cot A \cos A$

 

$=\sin A \tan A-\cos A \cot A$

Now,

$\frac{\sec A}{\operatorname{cosec}^{2} A}-\frac{\operatorname{cosec} A}{\sec ^{2} A}=\frac{\frac{1}{\cos A}}{\frac{1}{\sin ^{2} A}}-\frac{\frac{1}{\sin A}}{\frac{1}{\cos ^{2} A}}$

$=\frac{\sin ^{2} A}{\cos A}-\frac{\cos ^{2} A}{\sin A}$

$=\sin A \frac{\sin A}{\cos A}-\cos A \frac{\cos A}{\sin A}$

$=\sin A \tan A-\cos A \cot A$

Hence proved.