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

NDA 2015 · Mathematics · Inverse Trigonometric Functions

The value of (2 ^ -1 (1)/(5)-(π)/(4)) is

The value of tan⁡(2tan⁡−115−π4)\tan \left(2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right)is

Options

  1. A−717-\frac{7}{17}
  2. B5/16
  3. C5/4
  4. D717\frac{7}{17}

Correct answer

A. −717-\frac{7}{17}

Step-by-step solution

2tan⁡−1(15)=tan⁡−1[2×151−(15)2]=tan⁡−1[1024]=tan⁡−1(512)Let tan⁡(2tan⁡−115−π4)=x⇒tan⁡[tan⁡−1(512)−π4]=x⇒tan⁡−1(512)−π4=tan⁡−1x⇒tan⁡−1(512)−tan⁡−1(1)=tan⁡−1x⇒tan⁡−1[(512−1)1+(512)(1)]=tan⁡−1x⇒x=−7/1217/12=−7/17\begin{aligned} & 2 \tan ^{-1}\left(\frac{1}{5}\right)=\tan ^{-1}\left[\frac{2 \times \frac{1}{5}}{1-\left(\frac{1}{5}\right)^2}\right] \\ & =\tan ^{-1}\left[\frac{10}{24}\right] \\ & =\tan ^{-1}\left(\frac{5}{12}\right) \\ & \text{Let } \tan \left(2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right)=\mathrm{x} \\ & \Rightarrow \tan \left[\tan ^{-1}\left(\frac{5}{12}\right)-\frac{\pi}{4}\right]=\mathrm{x} \\ & \Rightarrow \tan { }^{-1}\left(\frac{5}{12}\right)-\frac{\pi}{4}=\tan ^{-1} \mathrm{x} \\ & \Rightarrow \tan ^{-1}\left(\frac{5}{12}\right)-\tan ^{-1}(1)=\tan ^{-1} \mathrm{x} \\ & \Rightarrow \tan ^{-1}\left[\frac{\left(\frac{5}{12}-1\right)}{1+\left(\frac{5}{12}\right)(1)}\right]=\tan ^{-1} \mathrm{x} \\ & \Rightarrow \quad \mathrm{x}=\frac{-7 / 12}{17 / 12}=-7 / 17\end{aligned}
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