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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
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B
1
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C
2
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D
3
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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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