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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 x = (2 ^ - 1 2 3 ) and y = ((1)/(2) ^ - 1 (12)/(5)), then

If x=sin⁡(2tan⁡−123)x = \sin \left(2\tan^{- 1} 2\sqrt{3}\right) and y=sin⁡(12tan⁡−1125),y = \sin \left(\dfrac{1}{2}\tan^{- 1} \dfrac{12}{5}\right), then

Options

  1. Ax=1−yx = 1 - y
  2. Bx2=1−2y{x}^{2} = 1 - 2y
  3. Cx2=1+y{x}^{2} = 1 + y
  4. Dy2=x3{y}^{2} = \dfrac{x}{\sqrt{3}}

Correct answer

D. y2=x3{y}^{2} = \dfrac{x}{\sqrt{3}}

Step-by-step solution

Let x=sin2\theta,\text{x} = \text{sin} 2\text{\theta}, where tan\theta=23\text{tan} \text{\theta} = 2\sqrt{3}
⇒x=2tan\theta1+tan2\theta=431+12=4313⇒ \text{x} = \dfrac{2\text{tan} \text{\theta}}{1 + {\text{tan}}^{2} \text{\theta}} = \dfrac{4\sqrt{3}}{1 + 12} = \dfrac{4\sqrt{3}}{13}
If \alpha=tan−1125,y=sin(\alpha2)=121−cos\alpha\text{\alpha} = {\text{tan}}^{- 1} \dfrac{12}{5}, \text{y} = \text{sin} \left(\dfrac{\text{\alpha}}{2}\right) = \dfrac{1}{\sqrt{2}}\sqrt{1 - \text{cos} \text{\alpha}}
=121−513=213= \dfrac{1}{\sqrt{2}}\sqrt{1 - \dfrac{5}{13}} = \dfrac{2}{\sqrt{13}}
∴y2=413⇒y2=x3∴ {\text{y}}^{2} = \dfrac{4}{13} ⇒ {\text{y}}^{2} = \dfrac{\text{x}}{\sqrt{3}}
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