If n is a positive integer and $I_n=\int \frac{Sin\left(nx\right)}{Cosx}dx$ then $I_n+I_{n-2}=$

$I_n=\int \frac{Sin\left(nx\right)}{Cosx}dx$

$I_{n-2}=\int \frac{Sin(n-2)x}{Cosx}dx$

$I_n+I_{n-2}=\int \frac{Sin nx+\sin \left(n-2\right)x}{Cosx}dx$

$$\begin{gathered} =\int \frac{2 \sin \left(n-1\right)x \cos x}{Cosx}dx \\ =\int 2 \sin \left(n-1\right)x dx=-2\frac{\cos \left(n-1\right)x}{n-1}+c \end{gathered}$$

$\left(\because \sin C+\sin D=2\sin \frac{{\displaystyle C+D}}{{\displaystyle 2}}.cos\frac{{\displaystyle C-D}}{{\displaystyle 2}}\right)$