Question

Please answer it.
it's urgent
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Answer 1

there is your answer

bas A ki place par theta h

Answer 2

$HEYA \: \\ \\ GIVEN \: \: QUESTION \: \: Is \: \: \\ \\ \frac{ \tan(x) }{1 + \sec(x) } - \frac{ \tan(x) }{1 - \sec(x) } = 2 \csc(x) \\ \\ lhs \\ \\ \frac{ \tan(x) }{1 + \sec(x) } - \frac{ \tan(x) }{1 - \sec(x) } \\ \\ replace \: \: \tan(x) \: by \: \frac{ \sin(x) }{ \cos(x) } \: \: \: and \: \\ replace \: \sec(x) \: \: by \: \: \frac{1}{ \cos(x) } \\ \\ \\ \frac{ \frac{ \sin(x) }{ \cos(x) } }{1 + \frac{1}{ \cos(x) } } \: - \: \frac{ \frac{ \sin(x) }{ \cos(x) } }{1 - \frac{1}{ \cos(x) } } \\ \\ \\ \frac{ \frac{ \sin(x) }{ \cos(x) } }{ \frac{ \cos(x) + 1}{ \cos(x) } } \: - \: \frac{ \frac{ \sin(x) }{ \cos(x) } }{ \frac{ \cos(x) - 1 }{ \cos(x) } } \\ \\ \\ \frac{ \sin(x) }{ \cos(x) + 1 } \: - \: \frac{ \sin(x) }{ \cos(x) - 1} \\ \\ \\ \\ \frac{ \sin(x) \cos(x) - \sin(x) - \sin(x ) \cos(x) - \sin(x) }{ \cos {}^{2} (x) - 1 {}^{2} } \\ \\ \\ \frac{ - 2 \sin(x) }{ - \sin {}^{2} (x) } \\ \\ \\ \frac{2}{ \sin(x) } \\ \\ = 2 \csc(x) \: \: \: hence \: \: proved \\ \\ \\ note \: \: \: \: \: \: \\ here \: \: \: a \: = x$