If 10 ^4 +15 ^4 =6, then the value of 27 cosec^6 +8 ^6 16 ^8 is

Question

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

Accepted Answer

$$\begin{aligned}
&\begin{aligned}
& 10 \sin ^4 \theta+15 \cos ^4 \theta=6 \
\Rightarrow & 10 \sin ^4 \theta+10 \cos ^4 \theta+5 \cos ^4 \theta=6 \
\Rightarrow & 10\left[\left(\sin ^2 \theta+\cos ^2 \theta\right)^2-2 \sin ^2 \theta \cos ^2 \theta\right]+5 \cos ^4 \theta=6 \
\Rightarrow & 10-20\left(1-\cos ^2 \theta\right) \cos ^2 \theta+5 \cos ^4 \theta=6
\end{aligned}\
\&\text { Let } \cos ^2 \theta=x\
\&10-20\left(x-x^2\right)+5 x^2=6
\end{aligned}$$

$$\begin{aligned}
\Rightarrow & 25 x^2-20 x+4=0 \
& (5 x-2)^2=0 \Rightarrow x=\frac{2}{5} \
\Rightarrow & \cos ^2 \theta=\frac{2}{5} \Rightarrow \sin ^2 \theta=\frac{3}{5}, \
& \sec ^2 \theta=\frac{5}{2}, \operatorname{cosec}^2 \theta=\frac{5}{3}
\end{aligned}$$

$$\begin{aligned}
\frac{27 \operatorname{cosec}^6 \theta+8 \sec ^6 \theta}{16 \sec ^8 \theta} & =\frac{27\left(\frac{5}{3}\right)^3+8\left(\frac{5}{2}\right)^3}{16\left(\frac{5}{2}\right)^4} \
& =\frac{5^3+5^3}{5^4}=\frac{2.5^3}{5^4}=\frac{2}{5}
\end{aligned}$$

Solutions · 1