Question

sen² x + cossec² x - sec² x . cossec² x=

Answer 1

$\sin^2x+cossec^2x-sec^2x\cdot cossec^2x=\\\\\sin^2x+\frac{1}{\sin^2x}-\frac{1}{\cos^2x}\cdot\frac{1}{\sin^2x}=\\\\\\\sin^2x+\frac{1}{\sin^2x}\left (1-\frac{1}{\cos^2x}\right )=\\\\\\\sin^2x+\frac{1}{\sin^2x}\left (\frac{\cos^2x-1}{\cos^2x}\right )=\\\\\\\sin^2x-\frac{1}{\sin^2x}\left (\frac{1-\cos^2x}{\cos^2x}\right )=\\\\\\\sin^2x-\frac{1}{\sin^2x}\left (\frac{\sin^2x}{\cos^2x}\right )=$

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