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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Advanced · Mathematics · Inverse Trigonometric Functions

Let S_n= ^ -1 ( 1 · ∑_ r=1 ^n (r-1) r), then:

Let Sn=tan⁡−1(sin⁡1⋅∑r=1nsec⁡(r−1)sec⁡r),then:S_n=\tan ^{-1} \left(\sin 1 \cdot \sum_{r=1}^n \sec (r-1) \sec r\right), then:

Options

  1. AS5=5−πS_5=5-\pi
  2. BS5=5−2πS_5=5-2 \pi
  3. CS5=10−3πS_5=10-3 \pi
  4. DS10=3π−10S_{10}=3 \pi-10

Correct answer

C. S5=10−3πS_5=10-3 \pi

Step-by-step solution

Sn=tan⁡−1(∑r=1nsin⁡(r−(r−1))cos⁡r⋅cos⁡(r−1))=tan⁡−1(∑r=1n(tan⁡r−tan⁡(r−1)))∴Sn=tan⁡−1(tan⁡(n))S_n=\tan ^{-1}\left(\sum_{r=1}^n \frac{\sin (r-(r-1))}{\cos r \cdot \cos (r-1)}\right)=\tan ^{-1}\left(\sum_{r=1}^n(\tan r-\tan (r-1))\right)\therefore \quad S_n=\tan ^{-1}(\tan (n))
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