If $\theta$ be acute angle and $\cos \theta=\frac{15}{17}$, then the value of $\cot \left(90^{\circ}-\theta\right)$ is

$\cos \theta=\frac{15}{17}$

$\therefore \quad \sin \theta=\sqrt{1-\cos ^{2} \theta}$

$=\sqrt{1-\left(\frac{15}{17}\right)^{2}}=\sqrt{1-\frac{225}{289}}$

$=\sqrt{\frac{289-225}{289}}=\sqrt{\frac{64}{289}}=\frac{8}{17}$

$\therefore \cot \left(90^{\circ}-\theta\right)=\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{\frac{8}{17}}{\frac{15}{17}}=\frac{8}{15}$