If I n = ∫ 0 π 4 tan n x d x , then 1 I 2 + 1 4 + 1 I 3 + I 5 + 1 I 4 + 1 6 =

1. $\frac{1}{I_9+I_{11}}$
2. $\frac{1}{I_{10}+I_{12}}$
3. $\frac{1}{I_{12}+I_{14}}$
4. $\frac{1}{I_{11}+I_{13}}$

Solution

Asked in: AP EAMCET 2022 (04 Jul Shift 2)

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