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Evaluate: (1 + sec A)/sec A = sin^2A/(1-cos A)

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Evaluate:

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

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

L.H.S. = (1 + sec A)/sec A

Since secant function is the inverse function of cos function and it is written as

= (1 + 1/cos A)/1/cos A

= (cos A + 1)/cos A/1/cos A

Therefore, (1 + sec A)/sec A = cos A + 1

R.H.S. = sin2A/(1-cos A)

We know that sin2A = (1 – cos2A), we get

= (1 – cos2A)/(1-cos A)

= (1-cos A)(1+cos A)/(1-cos A)

Therefore, sin2A/(1-cos A)= cos A + 1

L.H.S. = R.H.S.

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