Question

Assertion :If ${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{3\pi }{2}$,then $\frac{3{\sum }_{r=1}^{2008}(x^r+y^r)}{2{\sum }_{r=1}^{2008}\left(x^r{y}^r\right)}=3$ Reason: ${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{3\pi }{2}$ is possible only if $x=y=z=1$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion false but Reason is true

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

As $-1\le \sin \theta \le 1\Rightarrow -\frac{\pi }{2}\le {\sin }^{-1}\theta \le \frac{\pi }{2}$

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

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

Gives $x^r+y^r=1^r+1^r=2$ and $x^r.y^r=1^r.1^r=1$

Therefore,

$\frac{3{\sum }_{r=1}^{2008}(x^r+y^r)}{2{\sum }_{r=1}^{2008}\left(x^r{y}^r\right)}=\frac{\mathrm{3.2008.2}}{2.2008}=3$