Question

Sinx/cosecx-1 +cosx/1+secx=sinxcosx/sinx-cosx

Answer 1

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Answer 2

$\boxed{\huge{\mathfrak{Question}}}$

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$\frac{sin \: x}{cosec \: x - 1} \: + \frac{cos \: x}{1 + sec \: x} = \frac{sin \: x \: . \: cos \: x }{sin \: x - cos \: x}$

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$\boxed{\huge{\red{\mathfrak{Answer}}}}$

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$\boxed{\huge{\underline{\mathtt{L.H.S}}}}$

$\longrightarrow \: \frac{sin \: x}{ \frac{1 - sin \: x}{sin \: x} } + \frac{cos \: x}{ \frac{1 + cos \: x}{cos \: x} } \\ \\ \longrightarrow \: \frac{{sin}^{2} x}{1 - sin \: x} + \frac{ {cos}^{2}x }{1 + cos \: x} \\ \\ \longrightarrow \: \frac{ {sin}^{2}x(1 + cos \: x) + {cos}^{2} x(1 - sin \: x) }{(1 - sin \: x)(1 + cos \: x)} \\ \\ \longrightarrow \: \frac{ {sin}^{2}x + {sin}^{2}x \: . \: cos \: x + {cos}^{2}x - sin \: x \: .\: {cos}^{2}x }{(1 - sin \: x)(1 + cos \: x)} \\ \\ \longrightarrow \: \frac{1 + sin \: x \: . \: cos \: x(sin \: x - cos \: x)}{1 + cos \: x - sin \: x + sin \: x \: . \: cos \: x}$

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