Prove the following trigonometric identities.

Solution:

We have to prove $(\sec A-\operatorname{cosec} A)(1+\tan A+\cot A)=\tan A \sec A-\cot A \operatorname{cosec} A$

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

So,

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

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

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

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

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

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

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

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

$=\frac{-\cos ^{3} A+\sin ^{3} A}{\sin ^{2} A \cos ^{2} A}$

$=\frac{\sin ^{3} A-\cos ^{3} A}{\sin ^{2} A \cos ^{2} A}$

$=\frac{\sin ^{3} A}{\sin ^{2} A \cos ^{2} A}-\frac{\cos ^{3} A}{\sin ^{2} A \cos ^{2} A}$

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

$=\frac{\sin A}{\cos A} \frac{1}{\cos A}-\frac{\cos A}{\sin A} \frac{1}{\sin A}$

$=\tan A \sec A-\cot A \operatorname{cosec} A$

Hence proved.