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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 xx equal to?

Options

  1. Atan⁡−1(60119)\tan ^{-1} \left(\frac{60}{119}\right)
  2. Btan⁡−1(120119)\tan ^{-1} \left(\frac{120}{119}\right)
  3. Ctan⁡−1(90169)\tan ^{-1} \left(\frac{90}{169}\right)
  4. Dtan⁡−1(170169)\tan ^{-1} \left(\frac{170}{169}\right)

Correct answer

B. tan⁡−1(120119)\tan ^{-1} \left(\frac{120}{119}\right)

Step-by-step solution

x=4tan⁡−1(15)=2[2tan⁡−115]=2[tan⁡−1251−(15)2]=2tan⁡−1(25×2524)=2tan⁡−11024=2tan⁡−1512=tan⁡−1(2×5121−25144)=tan⁡−1120119.\begin{aligned} & x=4 \tan ^{-1}\left(\frac{1}{5}\right) \\ & =2\left[2 \tan ^{-1} \frac{1}{5}\right]=2\left[\tan ^{-1} \frac{\frac{2}{5}}{1-\left(\frac{1}{5}\right)^2}\right] \\ & =2 \tan ^{-1}\left(\frac{\frac{2}{5} \times 25}{24}\right)=2 \tan ^{-1} \frac{10}{24}=2 \tan ^{-1} \frac{5}{12} \\ & =\tan ^{-1}\left(\frac{2 \times \frac{5}{12}}{1-\frac{25}{144}}\right)=\tan ^{-1} \frac{120}{119}.\end{aligned}
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