Question

lf ${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{3\pi }{2}$ then $\sum \left(\frac{x^{201}+y^{201}}{x^{603}+y^{603}}\right)\left(\frac{x^{402}+y^{402}}{x^{804}+y^{804}}\right)$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D $3$

${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{3\pi }{2}$

$\Rightarrow {\sin }^{-1}x={\sin }^{-1}y={\sin }^{-1}z=\frac{\pi }{2}$
$\therefore x=y=z=1$
$\Sigma \left[\frac{x^{201}+y^{201}}{x^{603}+y^{603}}\right]\left[\frac{x^{402}+y^{402}}{x^{804}+y^{804}}\right]$
$=\left[\frac{x^{201}+y^{201}}{x^{603}+y^{603}}\right]\left[\frac{x^{402}+y^{402}}{x^{804}+y^{804}}\right]+\left[\frac{y^{201}+z^{201}}{y^{603}+z^{603}}\right]\left[\frac{y^{402}+z^{402}}{y^{804}+z^{804}}\right]+\left[\frac{x^{201}+z^{201}}{x^{603}+z^{603}}\right]\left[\frac{x^{402}+z^{402}}{x^{804}+z^{804}}\right]$
$=\left[\frac{1+1}{1+1}\right]\left[\frac{1+1}{1+1}\right]\times 3$
$=3$