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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 $L.H.S.=R.H.S.$
Solution:
$L.H.S.=( 1+cot A-cosecA)( 1+ tan A+secA)$
$=( 1+\frac{cosA}{sinA}-\frac{1}{sinA})( 1+\frac{sinA}{cosA}+\frac{1}{cosA})$
$=( \frac{sinA+cosA-1}{sinA})( \frac{cosA+sinA+1}{cosA})$
$=( \frac{( sinA+cosA-1)( sinA+cosA+1)}{sinAcosA})$
$=\frac{( sinA+cosA)^{2}-( 1)^{2}}{sinAcosA}$
$=\frac{sin^{2}A+cos^{2}A+2sinAcosA-1}{sinAcosA}$
$=\frac{1+2sinAcosA-1}{sinAcosA}$
$=\frac{2sinAcosA}{sinAcosA}$
$=2$
$R.H.S.$
Hence proved that $( 1+cot A-cosecA)( 1+ tan A+secA)=2$.
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