The number of solutions of the equation 2tan^ | vedclass

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(A) Given equation: $2	an^{-1}(\cos^2 x) = 	an^{-1}(2\csc^2 x)$.Using the formula $2	an^{-1} 	heta = 	an^{-1} \left( \frac{2	heta}{1 - 	heta^2}

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$m \le 1$

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(A) Given equation: $2	an^{-1}(\cos^2 x) = 	an^{-1}(2\csc^2 x)$.Using the formula $2	an^{-1} 	heta = 	an^{-1} \left( \frac{2	heta}{1 - 	heta^2} ight)$,we have:$	an^{-1} \left( \frac{2\cos^2 x}{1 - \cos^4 x} ight) = 	an^{-1} \left( \frac{2}{\sin^2 x} ight)$.Equating the arguments:$\frac{2\cos^2 x}{1 - \cos^4 x} = \frac{2}{\sin^2 x}$.Since $1 - \cos^4 x = (1 - \cos^2 x)(1 + \cos^2 x) = \sin^2 x(1 + \cos^2 x)$,the equation becomes:$\frac{2\cos^2 x}{\sin^2 x(1 + \cos^2 x)} = \frac{2}{\sin^2 x}$.Assuming $\sin^2 x 
eq 0$,we cancel $2/\sin^2 x$ from both sides:$\frac{\cos^2 x}{1 + \cos^2 x} = 1$.$\cos^2 x = 1 + \cos^2 x$,which implies $0 = 1$,a contradiction.Thus,there are no real values of $x$ that satisfy the equation.Therefore,$m = 0$.Since $m = 0$,the condition $m \le 1$ is satisfied.

The number of solutions of the equation $2\tan^{-1}(\cos^2 x) = \tan^{-1}(2\csc^2 x)$ in the interval $[0, 5\pi]$ is $m$. Then which of the following is true?

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