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

Given: $\frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A}$.

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

Solution:

$L.H.S.=\frac{1+\sec A}{\sec A}$

$=\frac{1}{secA}+\frac{secA}{secA}$   

$=cosA+1$           [$\because \frac{1}{secA}=cosA$ ]

$R.H.S.=\frac{\sin ^{2} A}{1-\cos A}$

$=\frac{1-cos^2A}{1-cosA}$     [$\because sin^2A=1-cos^2A$]

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

$=1+\cos A$

$=L.H.S.$

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Updated on: 10-Oct-2022

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