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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NDA 2019 · Mathematics · Inverse Trigonometric Functions

If ^ -1 (2 p)/(1+p^2)- ^ -1 (1-q^2)/(1+q^2)= ^ -1 (2 x)/(1-x^2) ,...

If sin⁡−12p1+p2−cos⁡−11−q21+q2=tan⁡−12x1−x2\sin ^{-1} \frac{2 p}{1+p^2}-\cos ^{-1} \frac{1-q^2}{1+q^2}=\tan ^{-1} \frac{2 x}{1-x^2}, then what is xx equal to ?

Options

  1. Ap+q1+pq\frac{p+q}{1+p q}
  2. Bp−q1+pq\frac{p-q}{1+p q}
  3. Cpq1+pq\frac{p q}{1+p q}
  4. Dp+q1−pq\frac{p+q}{1-p q}

Correct answer

B. p−q1+pq\frac{p-q}{1+p q}

Step-by-step solution

sin⁡−12p1+p2−cos⁡−11−q21+q2=tan⁡−12x1−x2\sin ^{-1} \frac{2 p}{1+p^2}-\cos ^{-1} \frac{1-q^2}{1+q^2}=\tan ^{-1} \frac{2 x}{1-x^2}⇒2tan⁡−1p−2tan⁡−1q=2tan⁡−1x⇒tan⁡−1p−tan⁡−1q=tan⁡−1x⇒tan⁡−1p−q1+pq=tan⁡−1x⇒x=p−q1+pq\begin{aligned} & \Rightarrow 2 \tan ^{-1} \mathrm{p}-2 \tan ^{-1} \mathrm{q}=2 \tan ^{-1} \mathrm{x} \\ & \Rightarrow \tan ^{-1} \mathrm{p}-\tan ^{-1} \mathrm{q}=\tan ^{-1} \mathrm{x} \\ & \Rightarrow \tan ^{-1} \frac{p-q}{1+p q}=\tan ^{-1} \mathrm{x} \\ & \Rightarrow \mathrm{x}=\frac{p-q}{1+p q}\end{aligned}$$
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