If ∫ 0 π / 2 tan n x d x = k ∫ 0 π / 2 cot n x d x , then

As, $I={\int }_0^{\pi /2}{\tan }^n\left(x\right)dx$

Apply the properties of definite integration: 

replace $x\to \frac{\mathrm{\pi }}{2}-t \Rightarrow \frac{dx}{dt}=-1$,

and upper limit $x=\frac{\mathrm{\pi }}{2} \Rightarrow \frac{\mathrm{\pi }}{2}-t=\frac{\mathrm{\pi }}{2} \Rightarrow t=0$
Similarly, lower limit $x=0 \Rightarrow \frac{\mathrm{\pi }}{2}-t=0 \Rightarrow t=\frac{\mathrm{\pi }}{2}$

$I={\int }_0^{\pi /2}{\tan }^n\left(x\right)dx={\int }_{\frac{\mathrm{\pi }}{2}}^0{\tan }^n\left(\frac{\mathrm{\pi }}{2}-t\right)\left(-dt\right)$$=-{\int }_0^{\frac{\mathrm{\pi }}{2}}{\tan }^n\left(\frac{\mathrm{\pi }}{2}-t\right)\left(-dt\right)={\int }_0^{\frac{\mathrm{\pi }}{2}}{\tan }^n\left(\frac{\mathrm{\pi }}{2}-t\right)\left(dt\right)$

$={\int }_0^{\frac{\mathrm{\pi }}{2}}{\cot }^n\left(t\right)dt$

Replacing $t\to x$, we get

$I={\int }_0^{\frac{\mathrm{\pi }}{2}}{\cot }^n\left(x\right)dx$
Hence, $k=1$