Question

tan theta by 1 minus cot theta + cot theta by 1 minus tan theta is equal to 1 + sin theta cos theta

Answer 1

Answer

Hope this is helpful to u

Answer 2

The RHS of the question you have typed is incorrect. My answer has the correct thing to prove in the RHS.

$\frac{ \tan(x) }{1 - \cot(x) } + \frac{ \cot(x) }{1 - \tan(x) } \\ = \frac{ \frac{ \sin(x) }{ \cos(x) } }{1 - \frac{ \cos(x) }{ \sin(x) } } + \frac{ \frac{ \cos(x) }{ \sin(x) } }{1 - \frac{ \sin(x) }{ \cos(x) } } \\ = \frac{ \frac{ \sin(x) }{ \cos(x) } }{ \frac{ \sin(x) - \cos(x) }{ \sin(x) } } + \frac{ \frac{ \cos(x) }{ \sin(x) } }{ \frac{ \cos(x) - \sin(x) }{ \cos(x) }} \\ = \frac{ \sin^{2} (x) }{ \cos(x)( \sin(x) - \cos(x)) } + \frac{ \cos ^{2} (x) }{ \sin(x)( \cos(x) - \sin(x) ) } \\ = \frac{ \sin^{2} (x) }{ \cos(x)( \sin(x) - \cos(x)) } - \frac{ \cos ^{2} (x) }{ \sin(x)( \sin(x) - \cos(x) ) } \\ = \frac{ \sin ^{3} (x) - \cos ^{3} (x) }{ \sin(x) \cos(x) ( \sin(x) - \cos(x)) } \\ = \frac{( \sin(x) - \cos(x))( \sin ^{2} (x) + \cos^{2} (x) + \sin(x) \cos(x)) }{ \sin(x) \cos(x)( \sin(x) - \cos(x)) } \\ = \frac{1 + \sin(x) \cos(x) }{ \sin(x) \cos(x) } \\ = \frac{1}{ \sin(x) \cos(x) } + 1 \\ = \boxed{\sec(x) \csc(x) + 1}$