Question

If $x=sin\left(2 t a n^{- 1} 3\right)$ and $y=sin\left(\frac{1}{2} \left(tan\right)^{- 1} \frac{4}{3}\right),$ then

• A. $2x=1-y$
• B. $x^{2}=1-2y$
• C. $x^{2}=1+y$
• D. $y^{2}=2x-1$

Answer 1

Answer: (D)

Let, $x=sin 2\theta $ (where $tan \theta =3$ )
$\Rightarrow x=\frac{2 tan \theta }{1 + t a n^{2} \theta }=\frac{6}{1 + 9}=\frac{3}{5}$
If $\alpha =tan^{- 1}\frac{4}{3}$
$\Rightarrow y=sin\left(\frac{\alpha }{2}\right)=\frac{1}{\sqrt{2}}\sqrt{1 - cos \alpha }$
$=\frac{1}{\sqrt{2}}\sqrt{1 - \frac{3}{5}}=\frac{1}{\sqrt{5}}$
$\therefore y^{2}=\frac{1}{5}\Rightarrow y^{2}=2x-1$