Question

1+seca/seca = sin^2a/1-cosa prove angles involved are acute for which the expressions r defined​

Answer 1

Answer:

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(1 + secA)/secA = sin²A/(1 – cosA)

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Solving R.H.S

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$\longrightarrow$ (1² – cos²A)/(1 – cosA)

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$\longrightarrow$ {(1 + cosA)(1 – cosA)}/(1 – cosA)

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$\longrightarrow$ 1 + cosA

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$\longrightarrow$ 1 + (1/secA)

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$\longrightarrow$ (secA + 1)/secA

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Hence proved.

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Remember:

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a² – b² = (a + b)(a – b)

sin²θ + cos²θ = 1