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JEE Advanced 2019Paper 2
Exam
JEE Advanced
Year
2019
Paper
Paper 2
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Advanced 2019 · Paper 2 · Mathematics · Trigonometric Ratios & Identities

The value of ^ - 1 ((1)/(4) ∑ _ k = 0 ^ 10 sec ((7π)/(12) +...

The value of \sec^{- 1} \left(\dfrac{1}{4}\mathop{∑}_{k = 0}^{10}sec\left(\dfrac{7\pi}{12} + \dfrac{k\pi}{2}\right)\sec \left(\dfrac{7\pi}{12} + \dfrac{\left(k + 1\right)\pi}{2}\right)\right)intheinterval[−π4,3π4]in the interval \left[- \dfrac{\pi}{4}, \dfrac{3\pi}{4}\right] equals

Correct answer

0

Step-by-step solution

sec−1[14∑k=010sec⁡(7π12+kπ2)sec⁡(7π12+kπ2+π2)]{\text{sec}}^{- 1} \left[\dfrac{1}{4} \mathop{∑}_{k = 0}^{10} \sec \left(\dfrac{7\pi}{12} + \dfrac{k\pi}{2}\right)\sec \left(\dfrac{7\pi}{12} + \dfrac{k\pi}{2} + \dfrac{\pi}{2}\right)\right]
{=sec−1[−14∑k=010sec⁡(7π12+kπ2)cosec(7π12+kπ2)](∴sec⁡(π2+θ)=−cosecθ)\begin{cases}= {\text{sec}}^{- 1} \left[\dfrac{- 1}{4} \mathop{∑}_{k = 0}^{10} \sec \left(\dfrac{7\pi}{12} + \dfrac{k\pi}{2}\right)\text{cosec}\left(\dfrac{7\pi}{12} + \dfrac{k\pi}{2}\right)\right] & \left(∴ \sec \left(\dfrac{\pi}{2} + \theta\right) = - \text{cosec}\theta\right)\end{cases}
=sec⁡−1(−14∑k=0101cos⁡(7π12+kπ2)sin⁡(7π12+kπ2))= \sec^{- 1}\left(- \dfrac{1}{4}\mathop{∑}_{k = 0}^{10}\dfrac{1}{\cos \left(\dfrac{7\pi}{12} + \dfrac{k\pi}{2}\right)\sin \left(\dfrac{7\pi}{12} + \dfrac{k\pi}{2}\right)}\right)
{=sec⁡−1(−14∑k=0102sin⁡(7π6+kπ))(∴2sinθcosθ=sin⁡2θ)\begin{cases}= \sec^{- 1} \left(- \dfrac{1}{4}\mathop{∑}_{k = 0}^{10}\dfrac{2}{\sin \left(\dfrac{7\pi}{6} + k\pi\right)}\right) & \left(∴ 2sin \theta cos \theta = \sin 2\theta\right)\end{cases}
now if {k=evensin⁡(2nπ+7π6)=sin7π6=−12\begin{cases}k = even & \sin \left(2n\pi + \dfrac{7\pi}{6}\right) = sin \dfrac{7\pi}{6} = - \dfrac{1}{2}\end{cases}
and if {k=oddsin⁡((2n+1)π+7π6)=−sin7π6=12\begin{cases}k = odd & \sin \left(\left(2n + 1\right)\pi + \dfrac{7\pi}{6}\right) = - sin \dfrac{7\pi}{6} = \dfrac{1}{2}\end{cases}
hence
=sec⁡−1(−12(1−12+112+1−12+…..))= \sec^{- 1} \left(- \dfrac{1}{2}\left(\dfrac{1}{- \dfrac{1}{2}} + \dfrac{1}{\dfrac{1}{2}} + \dfrac{1}{- \dfrac{1}{2}} + \ldots . .\right)\right)
=sec⁡−11= \sec^{- 1} 1
=0= 0
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