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

NDA 2015 · Mathematics · Inverse Trigonometric Functions

Consider the following : 1. ^ -1 (4)/(5)+ ^ -1 (3)/(5)=(π)/(2) 2. ^...

Consider the following : 1. sin⁡−145+sin⁡−135=π22.tan⁡−13+tan⁡−11=−tan⁡−1(2+3)\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{3}{5}=\frac{\pi}{2} 2. \tan ^{-1} \sqrt{3}+\tan ^{-1} 1=-\tan ^{-1}(2+\sqrt{3}) which of the above is/are correct?

Options

  1. A1 only
  2. B2 only
  3. CBoth 1 and 2
  4. DNeither 1 nor 2

Correct answer

C. Both 1 and 2

Step-by-step solution

sin⁡−1x+sin⁡−1y=sin⁡−1(x1−y2+y1−x2) If (−1≤x,y≤1)(x2+y2≤1)⇒sin⁡−1(45)+sin⁡−135=sin⁡−1[451−(35)2+351−(45)2]=sin⁡−1[1625+925]=sin⁡−1(1)=π2\begin{aligned} & \sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left(x \sqrt{1-y^2}+y \sqrt{1-x^2}\right) \\ & \text{ If }(-1 \leq x, y \leq 1) &\left(x^2+y^2 \leq 1\right) \\ & \Rightarrow \sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1} \frac{3}{5} \\ & =\sin ^{-1}\left[\frac{4}{5} \sqrt{1-\left(\frac{3}{5}\right)^2}+\frac{3}{5} \sqrt{1-\left(\frac{4}{5}\right)^2}\right] \\ & =\sin ^{-1}\left[\frac{16}{25}+\frac{9}{25}\right] \\ & =\sin ^{-1}(1)=\frac{\pi}{2} \end{aligned}\thereforeStatement(1)iscorrectAgain−Statement (1) is correct Again-tan⁡−1x+tan⁡−1y=π+tan⁡−1[x+y1−xy]If⁡;(x>0),(y>0) and  (xy>1)tan⁡−1(3)+tan⁡−1(1)=π+tan⁡−1[3+11−3]=π+tan⁡−1[(3+1)(1+3)(1−3)(1+3)]\begin{aligned} & \tan ^{-1} x+\tan ^{-1} y=\pi+\tan ^{-1}\left[\frac{x+y}{1-x y}\right] \\ & \operatorname{If} ;(x > 0),(y > 0) \text{\text{ and } }(x y > 1) \\ & \tan ^{-1}(\sqrt{3})+\tan ^{-1}(1)=\pi+\tan ^{-1}\left[\frac{\sqrt{3}+1}{1-\sqrt{3}}\right] \\ & =\pi+\tan ^{-1}\left[\frac{(\sqrt{3}+1)(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}\right] \end{aligned}$\begin{aligned} & =\pi+\tan ^{-1}\left(\frac{4+2 \sqrt{3}}{-2}\right) \\ & =\pi+\tan ^{-1}[-(2+\sqrt{3})] \\ & =\pi-\tan ^{-1}(2+\sqrt{3}) \quad \because \tan ^{-1}(-x)=-\tan ^{-1} x \end{aligned}\therefore$ Statement (1) is incorrect.
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