If 15 sin^{4} alpha + 10 cos^{4} alpha = 6 fo | vedclass

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(D) Given $15 \sin^{4} \alpha + 10 \cos^{4} \alpha = 6$.We know that $6 = 6(\sin^{2} \alpha + \cos^{2} \alpha)^{2} = 6(\sin^{4} \alpha + \cos^{4} \alp

Vedclass · Accepted Answer

$250$

Vedclass · Answer

(D) Given $15 \sin^{4} \alpha + 10 \cos^{4} \alpha = 6$.We know that $6 = 6(\sin^{2} \alpha + \cos^{2} \alpha)^{2} = 6(\sin^{4} \alpha + \cos^{4} \alpha + 2 \sin^{2} \alpha \cos^{2} \alpha)$.Substituting this into the equation:$15 \sin^{4} \alpha + 10 \cos^{4} \alpha = 6 \sin^{4} \alpha + 6 \cos^{4} \alpha + 12 \sin^{2} \alpha \cos^{2} \alpha$.$9 \sin^{4} \alpha + 4 \cos^{4} \alpha - 12 \sin^{2} \alpha \cos^{2} \alpha = 0$.$(3 \sin^{2} \alpha - 2 \cos^{2} \alpha)^{2} = 0$.This implies $3 \sin^{2} \alpha = 2 \cos^{2} \alpha$,so $	an^{2} \alpha = \frac{2}{3}$ and $\cot^{2} \alpha = \frac{3}{2}$.Now,$27 \sec^{6} \alpha + 8 \operatorname{cosec}^{6} \alpha = 27(1 + 	an^{2} \alpha)^{3} + 8(1 + \cot^{2} \alpha)^{3}$.$= 27(1 + \frac{2}{3})^{3} + 8(1 + \frac{3}{2})^{3}$.$= 27(\frac{5}{3})^{3} + 8(\frac{5}{2})^{3}$.$= 27 	imes \frac{125}{27} + 8 	imes \frac{125}{8}$.$= 125 + 125 = 250$.

If $15 \sin^{4} \alpha + 10 \cos^{4} \alpha = 6$ for some $\alpha \in R$,then the value of $27 \sec^{6} \alpha + 8 \operatorname{cosec}^{6} \alpha$ is equal to ....... .

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