If $\int \frac{1-7 \cos ^2 x}{\sin ^7 x \cos ^2 x} d x=\frac{f(x)}{\sin ^7 x}+C$ is equal to

$\tan x$

Let $I=\int \frac{1-7 \cos ^2 x}{\sin ^7 x \cos ^2 x} d x$

$\Rightarrow I=\int \frac{\sec ^2 x}{\sin ^7 x} d x-7 \int \frac{1}{\sin ^7 x} d x$

$\Rightarrow I=\int \frac{\tan x}{\sin ^7 x}-\int(-7)(\sin x)^{-8} \cos x \tan x d x-7 \int \frac{1}{\sin ^7 x} d x$

$\Rightarrow I=\int \frac{\tan x}{\sin ^7 x}+7 \int \frac{1}{\sin ^7 x} d x-7 \int \frac{1}{\sin ^7 x} d x+C$

$\Rightarrow I=\frac{\tan x}{\sin ^7 x}+C$

Hence, $f(x)=\tan x$.