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NDA 2013
Exam
NDA
Year
2013
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NDA 2013 · Mathematics · Inverse Trigonometric Functions

If ^ -1 ((2 a)/(1+a^2))+ ^ -1 ((2 b)/(1+b^2))=2 ^ -1 x , then x is...

If sin⁡−1(2a1+a2)+sin⁡−1(2b1+b2)=2tan⁡−1x\sin ^{-1}\left(\frac{2 a}{1+a^2}\right)+\sin ^{-1}\left(\frac{2 b}{1+b^2}\right)=2 \tan ^{-1} x, then xx is equal to

Options

  1. Aa−b1+ab\frac{a-b}{1+a b}
  2. Ba−b1−ab\frac{a-b}{1-a b}
  3. C2aba+b\frac{2 a b}{a+b}
  4. Da+b1−ab\frac{a+b}{1-a b}

Correct answer

D. a+b1−ab\frac{a+b}{1-a b}

Step-by-step solution

a>0,b>02tan⁡−1a+2tan⁡−1b=2tan⁡−1x2tan⁡−1(a+b1−ab)=2tan⁡−1x\begin{aligned} & a > 0, \quad b > 0 \\ & 2 \tan ^{-1} a+2 \tan ^{-1} b=2 \tan ^{-1} x \\ & 2 \tan ^{-1}\left(\frac{a+b}{1-a b}\right)=2 \tan ^{-1} x\end{aligned}x=\frac{a+b}{1-a b}$
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