If 15sin4[] + 10cos4[] = 6, for some [] R, then the value of 27sec6+ 8cosec6

Question

If 15sin4$$\alpha$$ + 10cos4$$\alpha$$ = 6, for some $$\alpha$$ $\in$R, then the value of

27sec6$\alpha$+ 8cosec6$\alpha$

Accepted Answer

$$
\begin{aligned}
& \text { Given, } 15 \sin ^4 \alpha+10 \cos ^4 \alpha=6 \\
& \Rightarrow \quad 15 \sin ^4 \alpha+10 \cos ^4 \alpha=6\left(\sin ^2 \alpha+\cos ^2 \alpha\right)^2 \\
& \Rightarrow \quad 15 \sin ^4 \alpha+10 \cos ^4 \alpha=6\left(\sin ^4 \alpha+\cos ^4 \alpha+2 \sin ^2 \alpha \cos ^2 \alpha\right) \\
& \Rightarrow 9 \sin ^4 \alpha+4 \cos ^4 \alpha-12 \sin ^2 \alpha \cos ^2 \alpha=0 \\
& \Rightarrow \quad\left(3 \sin ^2 \alpha-2 \cos ^2 \alpha\right)^2=0 \\
& \Rightarrow \quad 3 \sin ^2 \alpha-2 \cos ^2 \alpha=0 \\
& \Rightarrow \quad 3 \sin ^2 \alpha=2 \cos ^2 \alpha \\
& \Rightarrow \quad \tan ^2 \alpha=2 / 3 \\
& \therefore \quad \cot ^2 \alpha=3 / 2 \\
& \text { Now, } 27 \sec ^6 \alpha+8 \operatorname{cosec}^6 \alpha=27\left(\sec ^2 \alpha\right)^3+8\left(\operatorname{cosec}^2 \alpha\right)^3 \\
& =27\left(1+\tan ^2 \alpha\right)^3+8\left(+\cot ^2 \alpha\right)^3 \\
& =27\left(1+\frac{2}{3}\right)^3+8\left(1+\frac{3}{2}\right)^3=250
\end{aligned}
$$

Solutions · 1