If 10 sin^4 theta + 15 cos^4 theta = 6,then t | vedclass

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(A) Given $10 \sin^4 	heta + 15 \cos^4 	heta = 6$. Let $\sin^2 	heta = t$,then $\cos^2 	heta = 1 - t$. The equation becomes $10t^2 + 15(1 - t)^2 =

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$\frac{2}{5}$

Vedclass · Answer

(A) Given $10 \sin^4 	heta + 15 \cos^4 	heta = 6$. Let $\sin^2 	heta = t$,then $\cos^2 	heta = 1 - t$. The equation becomes $10t^2 + 15(1 - t)^2 = 6$. $10t^2 + 15(1 - 2t + t^2) = 6$. $10t^2 + 15 - 30t + 15t^2 = 6$. $25t^2 - 30t + 9 = 0$. $(5t - 3)^2 = 0$,so $t = \frac{3}{5}$. Thus,$\sin^2 	heta = \frac{3}{5}$ and $\cos^2 	heta = 1 - \frac{3}{5} = \frac{2}{5}$. Now,$\operatorname{cosec}^2 	heta = \frac{5}{3}$ and $\sec^2 	heta = \frac{5}{2}$. The expression is $\frac{27 \operatorname{cosec}^6 	heta + 8 \sec^6 	heta}{16 \sec^8 	heta} = \frac{27(\frac{5}{3})^3 + 8(\frac{5}{2})^3}{16(\frac{5}{2})^4}$. $= \frac{27 	imes \frac{125}{27} + 8 	imes \frac{125}{8}}{16 	imes \frac{625}{16}} = \frac{125 + 125}{625} = \frac{250}{625} = \frac{2}{5}$.

If $10 \sin^4 \theta + 15 \cos^4 \theta = 6$,then the value of $\frac{27 \operatorname{cosec}^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta}$ is:

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