The general value of theta satisfying the equ | vedclass

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(B) Given equation: $	an^2 	heta + \sec 2	heta = 1$.Using the identity $\sec 2	heta = \frac{1 + 	an^2 	heta}{1 - 	an^2 	heta}$,the equation be

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$m\pi, n\pi \pm \frac{\pi}{3}$

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(B) Given equation: $	an^2 	heta + \sec 2	heta = 1$.Using the identity $\sec 2	heta = \frac{1 + 	an^2 	heta}{1 - 	an^2 	heta}$,the equation becomes:$	an^2 	heta + \frac{1 + 	an^2 	heta}{1 - 	an^2 	heta} = 1$.Let $x = 	an^2 	heta$. Then $x + \frac{1+x}{1-x} = 1$.$x(1-x) + 1 + x = 1 - x$$x - x^2 + 1 + x = 1 - x$$3x - x^2 = 0 \Rightarrow x(3 - x) = 0$.So,$	an^2 	heta = 0$ or $	an^2 	heta = 3$.If $	an^2 	heta = 0$,then $	an 	heta = 0 \Rightarrow 	heta = m\pi$ for any integer $m$.If $	an^2 	heta = 3$,then $	an 	heta = \pm \sqrt{3} \Rightarrow 	heta = n\pi \pm \frac{\pi}{3}$ for any integer $n$.Thus,the general solution is $	heta = m\pi, n\pi \pm \frac{\pi}{3}$.

The general value of $\theta$ satisfying the equation $\tan^2 \theta + \sec 2\theta = 1$ is

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