Question

If ${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{3\pi }{2}$, then $\frac{{\sum }_{k=1}^2(x^{100k}+y^{106k})}{\sum x^{207}.y^{207}}$ is

• A. $\frac{1}{3}$
• B. $\frac{4}{3}$
• C. $\frac{2}{3}$
• D. None of these

Answer 1

Answer: (B)

$\frac{4}{3}$

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

It is possible only when

${\sin }^{-1}x=\frac{\pi }{2},{\sin }^{-1}y=\frac{\pi }{2}$ and ${\sin }^{-1}z=\frac{\pi }{2}$

[$\because \frac{-\pi }{2}\le {\sin }^{-1}x\le \frac{\pi }{2}$]

$\Rightarrow x=1,y=1$ and $z=1$

$\therefore \frac{{\sum }_{k=1}^2(x^{100k}+y^{106k})}{\sum x^{207}.y^{207}}=\frac{(x^{100}+y^{106})+(x^{200}+y^{212})}{x^{207}.y^{207}+y^{207}.z^{207}+z^{207}.x^{207}}$

$=\frac{1+1+1+1}{1+1+1}=\frac{4}{3}$