If 3(sec^2 theta + tan^2 theta) = 5,then the | vedclass

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(C) Given $3(\sec^2 	heta + 	an^2 	heta) = 5$,we have $\sec^2 	heta + 	an^2 	heta = \frac{5}{3}$.We know the identity $\sec^2 	heta - 	an^2

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$n\pi \pm \frac{\pi}{6}$

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(C) Given $3(\sec^2 	heta + 	an^2 	heta) = 5$,we have $\sec^2 	heta + 	an^2 	heta = \frac{5}{3}$.We know the identity $\sec^2 	heta - 	an^2 	heta = 1$.Subtracting the identity from the given equation: $(\sec^2 	heta + 	an^2 	heta) - (\sec^2 	heta - 	an^2 	heta) = \frac{5}{3} - 1$.$2 	an^2 	heta = \frac{2}{3} \Rightarrow 	an^2 	heta = \frac{1}{3}$.Since $	an^2 	heta = \frac{1}{3} = 	an^2(\frac{\pi}{6})$,the general solution for $	an^2 	heta = 	an^2 \alpha$ is $	heta = n\pi \pm \alpha$.Therefore,$	heta = n\pi \pm \frac{\pi}{6}$.

If $3(\sec^2 \theta + \tan^2 \theta) = 5$,then the general value of $\theta$ is

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