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Prove the following:$( i).\ ( \frac{\cot ^{2} A}{(\operatorname{cosec} A+1)^{2}}=\frac{1-\sin A}{1+\sin A})$$( ii).\ \frac{1+\sin A}{\cos A}+\frac{\cos A}{1+\sin A}=2 \sec A)$.
Given:
$( i).\ ( \frac{\cot ^{2} A}{(\operatorname{cosec} A+1)^{2}}=\frac{1-\sin A}{1+\sin A})$
$( ii).\ \frac{1+\sin A}{\cos A}+\frac{\cos A}{1+\sin A}=2 \sec A)$.
To do: To prove the given expressions.
Solution:
$( i)$. L.H.S.$=\frac{cot ^{2} A}{( cosec A+1)^{2}}$
$=\frac{cosec^2A-1}{( \frac{1}{sin A}+1)^2}$ [$\because cot^2A=cosec^2A-1$]
$=\frac{\frac{1}{sin^2A}-1}{( \frac{1+sinA}{sinA})^2}$ [$\because cosec^2A=\frac{1}{sin^2A}$]
$=\frac{\frac{1-sin^2A}{sin^2A}}{\frac{( 1+sinA)^2}{sin^2A}}$
$=\frac{1-sin^2A}{( 1+sinA)^2}$
$=\frac{( 1-sinA)( 1+sinA)}{( 1+sinA)^2}$
$=\frac{1-sinA}{1+sinA}$
$=R.H.S.$
Hence proved.
$( ii).$ LH.S.$= \frac{1+\sin A}{\cos A}+\frac{\cos A}{1+\sin A}=2 \sec A)$
$=\frac{( 1+sinA)^2+cos^2A}{cosA( 1+sinA)}$
$=\frac{1+sin^2A+2sinA+cos^2A}{cosA( 1+sinA)}$
$=\frac{1+( sin^2A+cos^2A)+2sinA}{cosA( 1+sinA)}$
$=\frac{1+1+2sinA}{cosA( 1+sinA)}$ [$\because sin^2A+cos^2A=1$]
$=\frac{2+2sinA}{cosA( 1+sinA)}$
$=\frac{2( 1+sinA)}{cosA( 1+sinA)}$
$=\frac{2}{cosA}$
$=2secA$ [$\because \frac{1}{cosA}=secA$]
$=R.H.S.$
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