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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NDA 2015 · Mathematics · Inverse Trigonometric Functions

Consider x=4 ^ -1 ((1)/(5)), y= ^ -1 ((1)/(70)) and z= ^ -1...

Consider x=4tan⁡−1(15),x=4 \tan ^{-1} \left(\frac{1}{5}\right),y=\tan ^{-1} \left(\frac{1}{70}\right)andandz=\tan ^{-1} \left(\frac{1}{99}\right)What is x−yx-y equal to?

Options

  1. Atan⁡−1(828845)\tan ^{-1} \left(\frac{828}{845}\right)
  2. Btan⁡−1(82878450)\tan ^{-1} \left(\frac{8287}{8450}\right)
  3. Ctan⁡−1(82818450)\tan ^{-1} \left(\frac{8281}{8450}\right)
  4. Dtan⁡−1(82878471)\tan ^{-1} \left(\frac{8287}{8471}\right)

Correct answer

C. tan⁡−1(82818450)\tan ^{-1} \left(\frac{8281}{8450}\right)

Step-by-step solution

x−y=tan⁡−1(120119)−tan⁡−1170=tan⁡−1∣120119−1701+(120119×170)∣=tan⁡−1∣8400−11983301+1208330∣\begin{aligned} & x-y=\tan ^{-1}\left(\frac{120}{119}\right)-\tan ^{-1} \frac{1}{70} \\ & =\tan ^{-1}\left|\frac{\frac{120}{119}-\frac{1}{70}}{1+\left(\frac{120}{119} \times \frac{1}{70}\right)}\right| \\ & =\tan ^{-1}\left|\frac{\frac{8400-119}{8330}}{1+\frac{120}{8330}}\right|\end{aligned}=\tan ^{-1}\left|\frac{\frac{8281}{8330}}{\frac{8450}{8330}}\right|=\tan ^{-1} \frac{8281}{8450}\therefore \quad$ option (C) is correct.
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