If frac{sin ^4 x}{2}+frac{cos ^4 x}{3}=frac{1 | vedclass

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(B) Given $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5}$.Let $\sin ^2 x = t$,then $\cos ^2 x = 1-t$. Since $0 \leq \sin ^2 x \leq 1$,we have $t

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$(A, C)$

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(B) Given $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5}$.Let $\sin ^2 x = t$,then $\cos ^2 x = 1-t$. Since $0 \leq \sin ^2 x \leq 1$,we have $t \in [0, 1]$.The equation becomes $\frac{t^2}{2} + \frac{(1-t)^2}{3} = \frac{1}{5}$.Multiplying by $30$ to clear denominators: $15t^2 + 10(1-2t+t^2) = 6$.$15t^2 + 10 - 20t + 10t^2 = 6$.$25t^2 - 20t + 4 = 0$.$(5t-2)^2 = 0$,which gives $t = \frac{2}{5}$.So,$\sin ^2 x = \frac{2}{5}$ and $\cos ^2 x = 1 - \frac{2}{5} = \frac{3}{5}$.Thus,$	an ^2 x = \frac{\sin ^2 x}{\cos ^2 x} = \frac{2/5}{3/5} = \frac{2}{3}$. This confirms $(A)$ is correct.Now check $(B)$: $\frac{\sin ^8 x}{8} + \frac{\cos ^8 x}{27} = \frac{(2/5)^4}{8} + \frac{(3/5)^4}{27} = \frac{16/625}{8} + \frac{81/625}{27} = \frac{2}{625} + \frac{3}{625} = \frac{5}{625} = \frac{1}{125}$. This confirms $(B)$ is correct.Therefore,the correct options are $(A)$ and $(B)$.

If $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5},$ then
$(A) \tan ^2 x=\frac{2}{3}$ $(B) \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}$
$(C) \tan ^2 x=\frac{1}{3}$ $(D) \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{2}{125}$

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If $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5},$ then$(A) \tan ^2 x=\frac{2}{3}$ $(B) \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}$$(C) \tan ^2 x=\frac{1}{3}$ $(D) \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{2}{125}$