Question

$\frac{ \sin( \alpha ) }{1 + \cos( \alpha ) } + \frac{ \sin( \alpha ) }{1 + \cos( \alpha ) } = 2 \csc( \alpha )$
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Answer 1

Correct Question:

To Prove :

${\sf{ {\dfrac{sin \alpha}{1 + cos \alpha}} + {\dfrac{1 + cos \alpha}{sin \alpha}} = 2 cosec \alpha}}$

Step-by-step explanation:

L.H.S. = ${\sf{ {\dfrac{sin \alpha}{1 + cos \alpha}} + {\dfrac{1 + cos \alpha}{sin \alpha}} }}$

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$\Longrightarrow{\sf{ {\dfrac{sin \alpha (sin \alpha) + 1 + cos \alpha (1 + cos \alpha) }{ (1 + cos \alpha)(sin \alpha)}}}}$

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$\Longrightarrow{\sf{ {\dfrac{ (sin \alpha)^2 + (1 + cos \alpha)^2}{ sin \alpha (1 + cos \alpha)}}}}$

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$\Longrightarrow{\sf{ {\dfrac{sin^2 \alpha + (1 + cos \alpha)^2}{ sin \alpha (1 + cos \alpha)}}}}$

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${\boxed{\tt{Identity \ : \ (a + b)^2 = a^2 + b^2 + 2ab}}}$

${\sf{Here, \ a = 1, \ b = cos \alpha}}$

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$\Longrightarrow{\sf{ {\dfrac{sin^2 \alpha + [ (1)^2 + (cos \alpha)^2 + 2(1)(cos \alpha) ] }{ sin \alpha (1 + cos \alpha)}}}}$

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$\Longrightarrow{\sf{ {\dfrac{sin^2 \alpha + [ 1 + cos^2 \alpha + 2 cos \alpha ]}{ sin \alpha (1 + cos \alpha)}}}}$

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Opening the bracket.

$\Longrightarrow{\sf{ {\dfrac{sin^2 \alpha + 1 + cos^2 \alpha + 2 cos \alpha}{sin \alpha (1 + cos \alpha)}}}}$

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${\boxed{\tt{Identity \ : \ sin^2 \theta + cos^2 \theta = 1}}}$

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$\Longrightarrow{\sf{ {\dfrac{1 + 1 + 2 cos \alpha}{sin \alpha (1 + cos \alpha)}}}}$

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$\Longrightarrow{\sf{ {\dfrac{2 + 2 cos \alpha}{sin \alpha (1 + cos \alpha)}}}}$

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$\Longrightarrow{\sf{ {\dfrac{2(1 + cos \alpha )}{sin \alpha (1 + cos \alpha)}}}}$

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$\Longrightarrow{\sf{ {\dfrac{2}{sin \alpha}}}}$

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We can write this as :

$\Longrightarrow{\sf{ 2 \times {\dfrac{1}{sin \alpha}}}}$

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${\boxed{\tt{Identity \ : \ {\dfrac{1}{sin \theta}} = cosec \theta}}}$

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$\Longrightarrow{\sf{2 cosec \alpha}}$

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= R.H.S.

Hence, proved !!