Mathematics · Trigonometry Ratios and Identities

JEE Advanced 2019 — Paper 2 — Question 33

\text{The value of } \sec^{-1} \left( \frac{1}{4} \sum_{k=0}^{10} \sec \left( \frac{7\pi}{12} + \frac{k\pi}{2} \right) \sec \left( \frac{7\pi}{12} + \frac{(k+1)\pi}{2} \right) \right) \text{ in the interval }

\left[ -\frac{\pi}{4}, \frac{3\pi}{4} \right] \text{ equals }

Answer: 0

Numerical answer — enter this value.

Step-by-step solution

\sec^{-1} \left( \frac{1}{4} \sum_{k=0}^{10} \sec \left( \frac{7\pi}{12} + \frac{k\pi}{2} \right) \sec \left( \frac{7\pi}{12} + \frac{(k+1)\pi}{2} \right) \right)

= \sec^{-1} \left( -\frac{1}{4} \sum_{k=0}^{10} \sec \left( \frac{7\pi}{12} + \frac{k\pi}{2} \right) \csc \left( \frac{7\pi}{12} + \frac{k\pi}{2} \right) \right)

= \sec^{-1} \left( -\frac{1}{2} \sum_{k=0}^{10} \frac{1}{\sin \left( \frac{7\pi}{6} \right) \cdot (-1)^k} \right)

= \sec^{-1} \left( -\frac{1}{2} \frac{1}{\sin \left( \frac{7\pi}{6} \right)} \right) = \sec^{-1} \left( -\frac{1}{2} \frac{1}{-1/2} \right) = \sec^{-1} (1) = 0.

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2019
Paper: Paper 2
Subject: Mathematics
Chapter: Trigonometry Ratios and Identities
Topic: Continued Sum or Product of Series of Trigonometric Ratios

The value of sec -1 ( 1/4 sum k=0 10 sec ( 7π/12 + kπ/2 ) sec ( 7π/12… | JEE Advanced 2019 PYQ with Solution · DhiX AI