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\(\frac{1 + secA}{secA} = \frac{sin^2A}{1 - cosA}\)

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in Trigonometry by (65.5k points)

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

\(\frac{1 + secA}{secA} = \frac{sin^2A}{1 - cosA}\)

[Hint: Simplify LHS and RHS separately]

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LHS = \(\frac{1 + secA}{secA} \)

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