Question

show that √1+sin A\1-sin A=sec A+tan A (0°<0<90°)​

Answer 1

Solution -

We have,

• $\sf{\sqrt{\dfrac{1 + sinA}{1 - sinA}} = secA + tanA}$

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Taking L.H.S

$\tt\longrightarrow{\sqrt{\dfrac{1 + sinA}{1 - sinA}}}$

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Rationalising the denominator

$\tt\longrightarrow{\sqrt{\dfrac{(1 + sinA)}{(1 - sinA)} \times \dfrac{(1 + sinA)}{(1 + sinA)}}}$

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$\tt\longrightarrow{\sqrt{\dfrac{(1 + sinA)^2}{1 - sin^2A}}}$

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$\tt\longrightarrow{\sqrt{\dfrac{(1 + sinA)^2}{cos^2A}}}$

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$\tt\longrightarrow{\dfrac{1 + sinA}{cosA}}$

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$\tt\longrightarrow{\dfrac{1}{cosA} + \dfrac{sinA}{cosA}}$

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$\tt\longrightarrow{secA + tanA}$

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$\tt\longrightarrow{R.H.S}$

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⇢ L.H.S = R.H.S

★ HENCE PROVED ★