Question

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

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

Answer 1

Answer: (D)

$y^2=1-x$

$y^2=si{n}^2\left(\frac{1}{2}ta{n}^{-1}\right(\frac{4}{3}\left)\right)$
$=\frac{1-cos\left(ta{n}^{-1}\right(\frac{4}{3}\left)\right)}{2}$
$=\frac{1-cos\left(co{s}^{-1}\right(\frac{3}{5}\left)\right)}{2}$
$y^2=\frac{1}{5}$ ...(i)
$x=sin\left(2ta{n}^{-1}\right(2\left)\right)$
$=2sin\left(ta{n}^{-1}\right(2\left)\right)co{s}^{-1}\left(ta{n}^{-1}\right(2\left)\right)$
$=2sin\left(si{n}^{-1}\right(\frac{2}{\sqrt{5}}\left)\right)co{s}^{-1}\left(cos\right(\frac{1}{\sqrt{5}}\left)\right)$
$=\frac{4}{5}$
$=1-y^2$
Hence
$x=1-y^2$
$y^2=1-x$