Explanation:
(D) : Given,\(I_{n}=\int_{0}^{\pi / 4} \tan ^{n} x d x\)
\(I_{n+2}=\int_{0}^{\pi / 4} \tan ^{n+2} x d x \tag{ii}\)
Adding equation (i) and (ii), we get -
\(I_{n}+I_{n+2} =\int_{0}^{\pi / 4}\left(\tan ^{n} x+\tan ^{n+2} x\right) d x\)
\(=\int_{0}^{\pi / 4} \tan ^{n} x\left(1+\tan ^{2} x\right) d x\)
\(=\int_{0}^{\pi / 4} \tan ^{n} x \times \sec ^{2} x d x\)
Put, \(\tan x=t\)
\(\sec ^{2} x d x=d t\)
\(\text { If, } \quad \mathrm{t}=0, \quad \tan \mathrm{x}=0\)
And, \(\mathrm{t}=\frac{\pi}{4}, \tan \mathrm{x}=1\)
\(\because \quad \mathrm{I}_{\mathrm{n}}+\mathrm{I}_{\mathrm{n}+2}=\int_{0}^{1}(\mathrm{t})^{\mathrm{n}} \mathrm{dt}\)
\(=\left[\frac{t^{n+1}}{n+1}\right]_{0}^{1}=\left[\frac{1}{n+1}-0\right]\)
\(I_{n}+I_{n+2}=\left(\frac{1}{n+1}\right)\)
If, \(\quad \mathrm{n}=8\)
So, \(\quad \mathrm{I}_{8}+\mathrm{I}_{10}=\frac{1}{8+1}=\frac{1}{9}\)