If $\int \frac{\sin ^4 x}{\cos ^8 x} d x=a \tan ^7 x+b \tan ^5 x+C$, then

$7 a=5 b$ 

We have,

$I=\int \frac{\sin ^4 x}{\cos ^8 x} d x$

$\Rightarrow I=\int \frac{\tan ^4 x}{\cos ^4 x} d x$       [Dividing $N^r$ and $D^r$ by $\cos ^4 x$]

$\Rightarrow I=\int \tan ^4 x \sec ^4 x d x$

$\Rightarrow I=\int \tan ^4 x\left(1+\tan ^2 x\right) d(\tan x)$

$\Rightarrow I=\int\left(\tan ^4 x+\tan ^6 x\right) d(\tan x)=\frac{\tan ^5 x}{5}+\frac{\tan ^7 x}{7}+C$

∴  $a=\frac{1}{7}$ and $b=\frac{1}{5} \Rightarrow 7 a=5 b$