Question

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

A
0
B
1
C
2
D
3

Solution

The correct option is D $$3$$$$\sin ^{ -1 }{ x } +\sin ^{ -1 }{ y } +\sin ^{ -1 }{ z } =\cfrac { 3\pi }{ 2 }$$$$\Rightarrow \sin ^{ -1 }{ x } =\sin ^{ -1 }{ y } =\sin ^{ -1 }{ z } =\cfrac { \pi }{ 2 }$$$$\therefore x=y=z=1$$$$\Sigma \left[ \cfrac { { x }^{ 201 }+{ y }^{ 201 } }{ { x }^{ 603 }+{ y }^{ 603 } } \right] \left[ \cfrac { { x }^{ 402 }+{ y }^{ 402 } }{ { x }^{ 804 }+{ y }^{ 804 } } \right]$$$$=\left[ \cfrac { { x }^{ 201 }+{ y }^{ 201 } }{ { x }^{ 603 }+{ y }^{ 603 } } \right] \left[ \cfrac { { x }^{ 402 }+{ y }^{ 402 } }{ { x }^{ 804 }+{ y }^{ 804 } } \right] +\left[ \cfrac { y^{ 201 }+{ z }^{ 201 } }{ { y }^{ 603 }+{ z }^{ 603 } } \right] \left[ \cfrac { { y }^{ 402 }+{ z }^{ 402 } }{ y^{ 804 }+{ z }^{ 804 } } \right] +\left[ \cfrac { { x }^{ 201 }+{ z }^{ 201 } }{ { x }^{ 603 }+{ z }^{ 603 } } \right] \left[ \cfrac { { x }^{ 402 }+{ z }^{ 402 } }{ { x }^{ 804 }+{ z }^{ 804 } } \right]$$$$=\left[ \cfrac { 1+1 }{ 1+1 } \right] \left[ \cfrac { 1+1 }{ 1+1 } \right] \times 3$$$$=3$$Mathematics

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