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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions

If ^ - 1 ( 2a 1 + a^ 2 ) + ^ - 1 ( 2b 1 + b^ 2 ) = 2 ^ - 1...

If sin⁡−1(2a1+a2)+sin⁡−1(2b1+b2)=\sin^{- 1} \left(\dfrac{2a}{1 + a^{2}}\right) + \sin^{- 1} \left(\dfrac{2b}{1 + b^{2}}\right) =2\cot^{- 1} \left(\dfrac{1}{x}\right), then xis equal to \left[\forall a, b \in \left(0, 1\right)\right]

Options

  1. Aa−b1+ab\dfrac{a - b}{1 + ab}
  2. Bb1+ab\dfrac{b}{1 + ab}
  3. Cb1−ab\dfrac{b}{1 - ab}
  4. Da+b1−ab\dfrac{a + b}{1 - ab}

Correct answer

D. a+b1−ab\dfrac{a + b}{1 - ab}

Step-by-step solution

Since, sin⁡−1(2x1+x2)=2tan⁡−1x\sin^{- 1} \left(\dfrac{2x}{1 + {x}^{2}}\right) = 2\tan^{- 1} x for x∈(−1,1)x \in \left(- 1, 1\right)
sin⁡−1(2a1+a2)+sin⁡−1(2b1+b2)=2cot⁡−1(1x)\sin^{- 1}\left(\dfrac{2a}{1 + {a}^{2}}\right) + \sin^{- 1}\left(\dfrac{2b}{1 + {b}^{2}}\right) = 2\cot^{- 1} \left(\dfrac{1}{x}\right)
⇒tan⁡−1a+tan⁡−1b=tan⁡−1x⇒ \tan^{- 1} a + \tan^{- 1} b = \tan^{- 1} x
⇒tan⁡−1(a+b1−ab)=tan⁡−1x⇒ \tan^{- 1} \left(\dfrac{a + b}{1 - ab}\right) = \tan^{- 1} x
⇒x=a+b1−ab⇒ x = \dfrac{a + b}{1 - ab}
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