Question

If $\frac{\sin x+\cos x}{\sin x-\cos x}+\frac{\sin x-\cos x}{\sin x+\cos x}=\frac{m{\sec }^{2}x}{{\tan }^{2}x-1}$. Find $m$

Answer 1

Consider $LHS$

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

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

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

Now consider $RHS$

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

$=\frac{m}{{\sin }^{2}x-{\cos }^{2}x}⟶\left(2\right)$

Equating (1) & (2)

$m=2$