The number of solutions of the equation sin5⁡x−cos5⁡x=1cos⁡x−1sin⁡x(sin⁡x≠cos⁡x) is

A
0
B
1
C
infinite
D
None of these

Solution:

$\begin{array}{ll} ∵ & \sin^{5} x - \cos^{5} x = \frac{\sin x - \cos x}{\sin xcos x} \\ \Rightarrow & \sin xcos x [\frac{\sin^{5} x - \cos^{5} x}{\sin x - \cos x}] = 1 \\ \Rightarrow & \frac{1}{2} \sin 2 x [\sin^{4} x + \sin^{3} xcos x + \sin^{2} {xcos}^{2} x + \sin {xcos}^{2} x + \cos^{4} x] = 1 \\ \begin{array}{ll} \Rightarrow & \sin 2 x [{(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} {xcos}^{2} x + \sin xcos x (\sin^{2} x + \cos^{2} x) + \sin^{2} {xcos}^{2} x] = 2 \end{array} \\ \Rightarrow & \sin 2 x [1 - \sin^{2} {xcos}^{2} x + \sin xcos x] = 2 \\ \Rightarrow & \sin^{3} 2 x - 2 \sin^{2} 2 x - 4 \sin 2 x + 8 = 0 \\ \Rightarrow & (\sin 2 x - 2)^{2} (\sin 2 x + 2) = 0 \end{array}$

$\Rightarrow \sin 2 x = \pm 2, which is not possible for any x$

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

Solution:

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