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

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(A) Given $\frac{\sin^4 x}{2} + \frac{\cos^4 x}{3} = \frac{1}{5}$.Multiplying by $6$,we get $3 \sin^4 x + 2 \cos^4 x = \frac{6}{5}$.$15 \sin^4 x + 10

Vedclass · Accepted Answer

$250$

Vedclass · Answer

(A) Given $\frac{\sin^4 x}{2} + \frac{\cos^4 x}{3} = \frac{1}{5}$.Multiplying by $6$,we get $3 \sin^4 x + 2 \cos^4 x = \frac{6}{5}$.$15 \sin^4 x + 10 \cos^4 x = 6$.Since $\cos^2 x = 1 - \sin^2 x$,we have $15 \sin^4 x + 10(1 - \sin^2 x)^2 = 6$.$15 \sin^4 x + 10(1 - 2 \sin^2 x + \sin^4 x) = 6$.$15 \sin^4 x + 10 - 20 \sin^2 x + 10 \sin^4 x = 6$.$25 \sin^4 x - 20 \sin^2 x + 4 = 0$.$(5 \sin^2 x - 2)^2 = 0$,which implies $\sin^2 x = \frac{2}{5}$.Then $\cos^2 x = 1 - \frac{2}{5} = \frac{3}{5}$.Thus,$\sec^2 x = \frac{5}{3}$ and $\operatorname{cosec}^2 x = \frac{5}{2}$.Now,$27 \sec^6 x + 8 \operatorname{cosec}^6 x = 27(\frac{5}{3})^3 + 8(\frac{5}{2})^3$.$= 27 	imes \frac{125}{27} + 8 	imes \frac{125}{8} = 125 + 125 = 250$.

If $\frac{\sin^4 x}{2} + \frac{\cos^4 x}{3} = \frac{1}{5}$,then $27 \sec^6 x + 8 \operatorname{cosec}^6 x = $

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