Question

$:\:\frac{1 + sinA - cosA}{1 + sinA + cosA}$= $\: = \: \: \sqrt{\frac{ 1 - cosA}{1 + cosA}}$​

Answer 1

Solve this
This is the answer

Answer 2

Solution -

$\tt\dashrightarrow{\dfrac{1+\sin\theta-\cos \theta}{1+\sin\theta+\cos \theta}= \sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}}}$

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Squaring both the sides

$\tt\dashrightarrow{\bigg( \dfrac{1+\sin\theta-\cos \theta}{1+\sin\theta+\cos \theta} \bigg)^2 = \bigg( \sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}} \bigg)^2}$

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$\tt\dashrightarrow{ \bigg( \dfrac{1+\sin\theta-\cos \theta}{1+\sin\theta+\cos \theta} \bigg)^2 = \dfrac{1-\cos\theta}{1+\cos\theta}}$

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Solving L.H.S

$\tt\dashrightarrow{\bigg( \dfrac{1+\sin\theta-\cos \theta}{1+\sin\theta+\cos \theta} \bigg)^2}$

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$\tt\dashrightarrow{\dfrac{1+1+2\sin\theta-2\sin\theta\cos\theta-2\cos\theta}{1+1+2\sin\theta+2\sin\theta\cos\theta+2\cos\theta}}$

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$\tt\dashrightarrow{\dfrac{2(1+\sin\theta)-2\cos\theta(1+\sin\theta)}{2(1+\sin\theta)+2\cos\theta(1+\sin\theta)}}$

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$\tt\dashrightarrow{\dfrac{(1+\sin\theta)(2-2\cos\theta)}{(1+\sin\theta)(2+2\cos\theta)}}$

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$\tt\dashrightarrow{\dfrac{2(1-\cos\theta)}{2(1+\cos\theta)}}$

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$\tt\dashrightarrow{\dfrac{(1-\cos\theta)}{(1+\cos\theta)}}$

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$\tt\dashrightarrow{L.H.S = R.H.S}$

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★ Hence proved ★

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