If ( ^4 x)/(5)+( ^4 x)/(4)=(1)/(9) , then which of the following...

We have $\frac{9}{5} \sin ^4 x+\frac{9}{4} \cos ^4 x=1\Rightarrow \frac{9}{5} \tan ^4 x+\frac{9}{4}=\sec ^4 x$ (on divide throughout by $\cos ^4 x$ ) $$\begin{aligned} & \Rightarrow \frac{9}{5} \tan ^4 x+\frac{9}{4}=\left(1+\tan ^2 x\right)^2 \\ & \Rightarrow 16 \tan ^4 x-40 \tan ^2 x+25=0 \Rightarrow\left(4 \tan ^2 x-5\right)^2=0\end{aligned}$$\therefore \quad \tan ^2 x=\frac{5}{4}$Now,$\sec ^2 x=1+\tan ^2 x=1+\frac{5}{4}=\frac{9}{4}$and$\operatorname{cosec}^2 x=1+\cot ^2 x=1+\frac{4}{5}=\frac{9}{5}$ Hence 64 \sec ^6 x+125 \operatorname{cosec}^6 x=\left(64 \times \frac{729}{64}\right)+\left(125 \times \frac{729}{125}\right)=1458