Question

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

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A).
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
• C. (A) is true but (R) is false.
• D. (A) is false but (R) is true.

Answer 1

Answer: (A)

$x=y=z=1$
$\therefore x^{2008}+y^{2008}=2$& $x^{2008}{y}^{2008}=1$
$\therefore \frac{3\sum _{r=1}^{2008}(x^{2008r}+y^{2008r})}{2008\sum x^{2008}\cdot y^{2008}}$
$=\frac{3\times 2008\times 2}{2008\times 3}=2$