Question

Prove that $[\sin x+\sec x{]}^2+[\cos x+\csc x{]}^2=[1+\sec x.\csc x{]}^2$

Answer 1

Simplifying the LHS of ${\left[\sin x+\sec x\right]}^2+{\left[\cos x+\csc x\right]}^2={\left[1+\sec x\cdot \csc x\right]}^2$

${\left[\sin x+\sec x\right]}^2+{\left[\cos x+\csc x\right]}^2={\left[\frac{\sin x\cos x+1}{\cos x}\right]}^2+{\left[\frac{\sin x\cos x+1}{\sin x}\right]}^2$

$=\frac{{\sin }^{2}x{\cos }^{2}x+1+2\sin x\cos x}{{\cos }^{2}x}+\frac{{\sin }^{2}x{\cos }^{2}x+1+2\sin x\cos x}{{\sin }^{2}x}$

$=\frac{{\sin }^{4}x{\cos }^{2}x+{\sin }^{2}x+2{\sin }^{3}x\cos x+{\sin }^{2}x{\cos }^{4}x+{\cos }^{2}x+2\sin x{\cos }^{3}x}{{\cos }^{2}x{\sin }^{2}x}$

$=\frac{\left({\sin }^{2}x+{\cos }^{2}x\right)+\left({\sin }^{4}x{\cos }^{2}x+{\sin }^{2}x{\cos }^{4}x\right)+\left(2{\sin }^{3}x\cos x+2\sin x{\cos }^{3}x\right)}{{\cos }^{2}x{\sin }^{2}x}$

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

$=\frac{1+{\sin }^{2}x{\cos }^{2}x+2\sin x\cos x}{{\cos }^{2}x{\sin }^{2}x}$

$=\frac{1}{{\cos }^{2}x{\sin }^{2}x}+\frac{{\cos }^{2}x{\sin }^{2}x}{{\cos }^{2}x{\sin }^{2}x}+\frac{2\sin x\cos x}{{\cos }^{2}x{\sin }^{2}x}$

$={\sec }^{2}x{\csc }^{2}x+1+2\sec x\csc x$

$={\left[1+\sec x\csc x\right]}^2$

$=RHS$