Question: If $\mathrm{I}_{\mathrm{n}}=\int_{\pi / 4}^{\pi / 2} \cot ^{\mathrm{n}} \mathrm{xdx}$, then:
(1) $\frac{1}{\mathrm{I}_{2}+\mathrm{L}_{4}}, \frac{1}{\mathrm{I}_{3}+\mathrm{I}_{5}}, \frac{1}{\mathrm{~L}_{4}+\mathrm{I}_{6}}$ are in G.P.
(2) $\frac{1}{\mathrm{I}_{2}+\mathrm{I}_{1}}, \frac{1}{\mathrm{I}_{3}+\mathrm{I}_{6}}, \frac{1}{\mathrm{I}_{1}+\mathrm{I}_{6}}$ are in A.P.
(3) $\mathrm{I}_{2}+\mathrm{I}_{4}, \mathrm{I}_{3}+\mathrm{I}_{5}, \mathrm{I}_{4}+\mathrm{I}_{6}$ are in A.P.
(4) $\mathrm{I}_{2}+\mathrm{I}_{4},\left(\mathrm{I}_{3}+\mathrm{I}_{5}\right)^{2}, \mathrm{I}_{4}+\mathrm{I}_{6}$ are in G.P.
Correct Option: , 2
Solution:
$I_{n+2}+I_{n}=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot ^{n} x \cdot \cos e c^{2} x d x=\left[\frac{-(\cot x)^{n+1}}{n+1}\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$
$I_{n+2}+I_{n}=\frac{1}{n+1}$
$I_{2}+I_{4}=\frac{1}{3}, I_{3}+I_{5}=\frac{1}{4}, I_{4}+I_{6}=\frac{1}{5}$
