Prove that: $( 1+cot A-cosecA)( 1+ tan A+secA)=2$

Given: $( 1+cot A-cosecA)( 1+ tan A+secA)=2$

To do: To prove that $L.H.S.=R.H.S.$

$L.H.S.=( 1+cot A-cosecA)( 1+ tan A+secA)$ 

$=( 1+\frac{cosA}{sinA}-\frac{1}{sinA})( 1+\frac{sinA}{cosA}+\frac{1}{cosA})$ 

$=( \frac{sinA+cosA-1}{sinA})( \frac{cosA+sinA+1}{cosA})$

$=( \frac{( sinA+cosA-1)( sinA+cosA+1)}{sinAcosA})$

$=\frac{( sinA+cosA)^{2}-( 1)^{2}}{sinAcosA}$

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

$=\frac{1+2sinAcosA-1}{sinAcosA}$

$=\frac{2sinAcosA}{sinAcosA}$

$=2$

$R.H.S.$

Hence proved that  $( 1+cot A-cosecA)( 1+ tan A+secA)=2$.