Question

The number of solutions of the equation ${\sin }^{5}x-{\cos }^{5}x=\frac{1}{\cos x}-\frac{1}{\sin x}(\sin x\ne \cos x)$ is

• A. 0
• B. 1
• C. $\infty$
• D. none of these

Answer 1

Answer: (A)

Given equation is ${\sin }^{5}x-{\cos }^{5}x=\frac{1}{\cos x}-\frac{1}{\sin x}$
$(\sin x-\cos x)[{\sin }^{4}x+{\sin }^{3}x\cos x+{\sin }^{2}x{\cos }^{2}x+\sin$ $x{\cos }^{3}x+{\sin }^{4}x]=\frac{\sin x-\cos x}{\sin x\cos x}$
$\Rightarrow \left({\sin }^{4}x+{\cos }^{4}x\right)+{\sin }^{3}x\cos x+\sin x{\cos }^{3}x+{\sin }^{2}x$
${\cos }^{2}x=\frac{1}{\sin x\cos x}$
$\Rightarrow 1-2{\sin }^{2}x{\cos }^{2}x+\sin x\cos x\left[{\sin }^{2}x+{\cos }^{2}x\right]+{\sin }^2$
$x{\cos }^{2}x=\frac{1}{\sin x\cos x}$
$\Rightarrow 1-{\sin }^{2}x{\cos }^{2}x+\sin x\cos x=\frac{1}{\sin x\cos x}$
$\Rightarrow 1-{\sin }^{2}x{\cos }^{2}x=\frac{1}{\sin x\cos x}-\sin x\cos x$
$\Rightarrow 1=\frac{1}{\sin x\cos x}$
$\sin (2x)=2$ which is impossible
$\therefore$ No solution