If In=∫tann⁡xdx then (n−1)In+In−2 is equal to

Here,In=∫tann⁡xdx=∫tann−2⁡xtan2⁡xdx=∫tann−2⁡sec2⁡x−1dx=∫tann−2⁡x⋅sec2⁡xdx−∫tann−2⁡xdx=∫tann−2⁡xsec2⁡xdx−In−2In+In−2=tann−1⁡xn−1Hence, (n−1)In+In−2=tann−1⁡x