Question

P.T:
[Cos x ÷(1-tan x)]+[sin x ÷(1-cot x)]= sin x+cos x

Answer 1

$\frac{ \cos(x) }{1 - \tan(x) } + \frac{ \sin(x ) }{1 - \cot(x) } \\ \frac{ \cos(x) }{1 - \frac{ \sin(x) }{ \cos(x) } } + \frac{ \sin(x) }{1 - \frac{ \cos(x) }{ \sin(x) } } \\ \frac{ \cos(x) }{ \ \frac{ \cos(x) - x \sin(x) }{ \cos(x) } } + \frac{ \sin(x) }{ \frac{ \sin(x) - \cos(x) }{ \sin(x ) } } \\ \frac{ \cos { }^{2} (x) }{ \cos(x) - \sin(x) } + \frac{ \sin {}^{2} ( { {x}^{} }^{} ) }{ \sin(x) - \cos(x) } \\ \ \ \frac{ \cos {}^{2} (x) }{ \cos(x) - \sin(x) } - \frac{ \sin {}^{2} (x) }{ \cos(x) - \sin(x) } \\ \frac{ \ \cos {}^{2} (x) - \sin {}^{2} (x) }{ \cos(x) - \sin(x) } \\ \frac{( \cos(x) + \sin(x) )( \cos(x) - \sin(x) ) }{ \cos(x) - \sin(x) } \\ \ = cos(x) + \sin(x)$



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