JEE Advanced 2019 · Paper 2, Mathematics question 12

Simplify one term.
$$\sec A\sec\left(A+\frac\pi2\right)=-\frac1{\sin A\cos A}=-\frac2{\sin2A}.$$
With $A_k=\frac{7\pi}{12}+\frac{k\pi}2$:
$$\sin2A_k=\sin\left(\frac{7\pi}6+k\pi\right)=(-1)^{k+1}\cdot\frac12,$$
so each term is $4(-1)^k$.

Sum. For $k=0,\dots,10$ there are six $+4$ terms and five $-4$ terms, giving 4. Then $\frac14\cdot4=1$, and $\sec^{-1}1=0$.

Answer: 0