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  4. The number of solutions of the equation sin 5 x- cos 5 x=(1/ cos x)-(1/ sin x)( sin x ≠ cos x) is

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

Trigonometric Functions

Solution:

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