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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NDA 2016 · Mathematics · Inverse Trigonometric Functions

Consider the following statements: 1. There exists θ ∈(-(π)/(2),...

Consider the following statements: 1. There exists θ∈(−π2,π2)\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) for which tan⁡−1(tan⁡θ)≠θ\tan ^{-1}(\tan \theta) \neq \theta. 2. sin⁡−1(13)−sin⁡−1(15)=sin⁡−1(22(3−1)15)\quad \sin ^{-1}\left(\frac{1}{3}\right)-\sin ^{-1}\left(\frac{1}{5}\right)=\sin ^{-1}\left(\frac{2 \sqrt{2}(\sqrt{3}-1)}{15}\right) Which of the above statements is/are correct?

Options

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

Correct answer

B. 2 only

Step-by-step solution

θ∈(−π2,π2) for  which tan⁡−1(tan⁡θ)=θ\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \text{\text{ for } which } \tan ^{-1}(\tan \theta)=\theta Hence, statement (1) is incorrect.  if x≤1;y≤1x2+y2≤1∴sin⁡−1(x)−sin⁡−1y=sin⁡−1[x1−y2−y1−x2]\begin{aligned} & \text{ if } x \leq 1 ; y \leq 1 & x^2+y^2 \leq 1 \\ & \therefore \sin ^{-1}(x)-\sin ^{-1} \mathrm{y}=\sin ^{-1}\left[\mathrm{x} \sqrt{1-\mathrm{y}^2}-\mathrm{y} \sqrt{1-\mathrm{x}^2}\right] \end{aligned}⇒sin⁡−113−sin⁡−115=sin⁡−1[131−125−151−19]=sin⁡−1[13×524−15×38]=sin⁡−1[8×3−815]=sin⁡−1[22(3−1)15]\begin{aligned} \Rightarrow & \sin ^{-1} \frac{1}{3}-\sin ^{-1} \frac{1}{5} \\ \quad & =\sin ^{-1}\left[\frac{1}{3} \sqrt{1-\frac{1}{25}}-\frac{1}{5} \sqrt{1-\frac{1}{9}}\right] \\ & =\sin ^{-1}\left[\frac{1}{3 \times 5} \sqrt{24}-\frac{1}{5 \times 3} \sqrt{8}\right] \\ & =\sin ^{-1}\left[\frac{\sqrt{8 \times 3}-\sqrt{8}}{15}\right]=\sin ^{-1}\left[\frac{2 \sqrt{2}(\sqrt{3}-1)}{15}\right] \end{aligned}$ Hence, statement (2) is correct.
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