Question

I don't know how to solve equations like this

Answer 1

$\frac{1}{1+ \tan^{2}x } + \frac{1}{1+ \cot^{2}x } \\ = \frac{1}{\sec^{2}x} + \frac{1}{\csc^{2}x} \ \ \ \ \ \ \ [1+ \tan^{2}x = \sec^{2}x \ and \ 1+ \cot^{2}x = \csc^{2}x] \\ = \frac{\csc^{2}x+\sec^{2}x}{(\sec^{2}x)(\csc^{2}x)} \\ = \frac{ \frac{1}{\sin^{2}x}+ \frac{1}{\cos^{2}x}}{ \frac{1}{\sin^{2}x}\times \frac{1}{\cos^{2}x}} \\ = \frac{ \frac{\cos^{2}x+\sin^{2}x}{(\sin^{2}x)(\cos^{2}x)} }{ \frac{1}{(\sin^{2}x)(\cos^{2}x)}} \\ = \cos^{2}x+\sin^{2}x \\ =1$