Answer:
We have, \[{{I}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}x\,dx,}\] \[\Rightarrow \] \[{{I}_{n\,+\,2}}=\int_{0}^{\pi /4}{{{\tan }^{n\,+\,2}}x\,dx}\] \[\Rightarrow \] \[\,{{I}_{n}}\,+\,{{I}_{n\,+\,2}}=\int_{0}^{\pi /4}{{{\tan }^{n}}x\,dx}+\int_{0}^{\pi /4}{{{\tan }^{n\,+\,2}}x\,dx}\] \[=\int_{0}^{\pi /4}{{{\tan }^{n}}x\,(1+{{\tan }^{2}}x)\,dx}\] \[=\int_{0}^{\pi /4}{{{\tan }^{n}}x{{\sec }^{2}}x\,dx}\] Put \[\tan \,x=t\] \[\Rightarrow \] \[{{\sec }^{2}}x\,dx=dt\] Also, when x = 0, then t = 0 And when \[x=\frac{\pi }{4},\] then t = 1 \[\therefore \] \[{{I}_{n}}+{{I}_{n\,+\,2}}=\int_{0}^{1}{{{t}^{n}}}\,dt=\left[ \frac{{{t}^{n\,+\,1}}}{n+1} \right]_{0}^{1}=\frac{1}{n+1}\] Hence proved.
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