Prove the following trigonometric identities.

We have to prove $(1+\cot A-\operatorname{cosec} A)(1+\tan A+\sec A)=2$

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

So,

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

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

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

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

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

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

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

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

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

$=2$

Hence proved.