The general solutions of the equation tan^2 t | vedclass

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(A) Given equation: $	an^2 	heta + \sec 2	heta = 1$We know that $	an^2 	heta = \frac{1 - \cos 2	heta}{1 + \cos 2	heta}$ and $\sec 2	heta = \fr

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$n\pi, n\pi \pm \frac{\pi}{3}, n \in \mathbb{Z}$

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(A) Given equation: $	an^2 	heta + \sec 2	heta = 1$We know that $	an^2 	heta = \frac{1 - \cos 2	heta}{1 + \cos 2	heta}$ and $\sec 2	heta = \frac{1}{\cos 2	heta}$.Substituting these,we get $\frac{1 - \cos 2	heta}{1 + \cos 2	heta} + \frac{1}{\cos 2	heta} = 1$.Let $x = \cos 2	heta$. Then $\frac{1-x}{1+x} + \frac{1}{x} = 1$.$\frac{x(1-x) + 1+x}{x(1+x)} = 1 \implies x - x^2 + 1 + x = x + x^2$.$2x^2 - x - 1 = 0 \implies (2x + 1)(x - 1) = 0$.Case $1$: $\cos 2	heta = 1 \implies 2	heta = 2n\pi \implies 	heta = n\pi$.Case $2$: $\cos 2	heta = -\frac{1}{2} \implies 2	heta = 2n\pi \pm \frac{2\pi}{3} \implies 	heta = n\pi \pm \frac{\pi}{3}$.

The general solutions of the equation $\tan^2 \theta + \sec 2\theta = 1$ are

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