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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 (1)/(3) + ^ -1 (2)/(3) = ^ -1 x , then the value of x is

If sin⁡−113+sin⁡−123=sin⁡−1x\sin^{-1} \dfrac{1}{3} + \sin^{-1} \dfrac{2}{3} = \sin^{-1} x, then the value of xx is

Options

  1. A00
  2. B(5−42)9\dfrac{( \sqrt{5} - 4\sqrt{2} )}{9}
  3. C(5+42)9\dfrac{( \sqrt{5} + 4\sqrt{2} )}{9}
  4. Dπ2\dfrac{\pi}{2}

Correct answer

C. (5+42)9\dfrac{( \sqrt{5} + 4\sqrt{2} )}{9}

Step-by-step solution

Given that, sin⁡−113+sin⁡−123=sin⁡1x\sin^{- 1} \dfrac{1}{3} + \sin^{- 1} \dfrac{2}{3} = \sin^{1} x

Taking sine on both sides

⇒(131−49+231−19)=x⇒ \left(\dfrac{1}{3}\sqrt{1 - \dfrac{4}{9}} + \dfrac{2}{3}\sqrt{1 - \dfrac{1}{9}}\right) = x

⇒(13.53+23.83)=x⇒ \left(\dfrac{1}{3} . \dfrac{\sqrt{5}}{3} + \dfrac{2}{3} . \dfrac{\sqrt{8}}{3}\right) = x

⇒(5+429)=x⇒ \left(\dfrac{\sqrt{5} + 4\sqrt{2}}{9}\right) = x

$∴ x = \left(\dfrac{\sqrt{5} + 4\sqrt{2}}{9}\right)
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