Question

Prove that:$(1+\cot A-\text{cosec}A)(1+\tan A+\sec A)=2.$

Answer 1

LHS =$1+\cot A-\text{cosec}A=1+\frac{\cos A}{\sin A}-\frac{1}{\sin A}=\frac{(\sin A+\cos A-1)}{\sin A}$

$1+\tan A+\text{sec}A=1+\frac{\sin A}{\cos A}+\frac{1}{\cos A}=\frac{(\sin A+\cos A+1)}{\cos A}$

$(1+\cot A-\text{cosec}A)(1+\tan A+\text{sec}A)$

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

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

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

Hence Proved