Question

$\frac{1}{\sec x-\tan x}-\frac{1}{\cos x}=\frac{1}{\cos x}-\frac{1}{\sec x+\tan x}$

Answer 1

$LHS=\left(\frac{1}{secx-tanx}\right)-\left(\frac{1}{cosx}\right)=\left(\frac{se{c}^{2}x-ta{n}^{2}x}{secx-tanx}\right)-\left(\frac{1}{cosx}\right)=secx+tanx-secx=\left(\frac{1}{cosx}\right)-(secx-tanx)=\left(\frac{1}{cosx}\right)-\left(\frac{secx-tanx}{se{c}^{2}x-ta{n}^{2}x}\right)=\left(\frac{1}{cosx}\right)-\left(\frac{1}{secx+tanx}\right)=RHS$