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If {\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \frac{{3\pi }}{2} then value of \sum {\frac{{\left( {{x^{101}} + {y^{101}}} \right)\left( {{x^{505}} + {y^{505}}} \right)}}{{\left( {{x^{303}} + {y^{303}}} \right)\left( {{x^{606}} + {y^{606}}} \right)}}}

Option: 1 1

Option: 2 2

Option: 3 3

Option: 4 4

Answers (1)

best_answer

As we learnt

 

Formulae of Inverse Trigonometric Functions -

 

\sin ^{-1}x + \sin ^{-1}y= \sin ^{-1}\left [ x\sqrt{1-y^{2}}+ y \sqrt{1-x^{2}} \right ]

\sin ^{-1}x + \sin ^{-1}y=\pi - \sin ^{-1}\left [ x\sqrt{1-y^{2}}+ y \sqrt{1-x^{2}} \right ]

- wherein

x\geqslant 0, y\geqslant 0, x^{2}+ y^{2}\leqslant 1

 

x\geqslant 0, y\geqslant 0, x^{2}+y^{2}> 1

 

 

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

Is only possible if x=y=z=1 then {\sin ^{ - 1}}x = {\sin ^{ - 1}}y = {\sin ^{ - 1}}z = \frac{\pi }{2}

 \sum {\frac{{({x^{101}} + {y^{101}})({x^{505}} + {y^{505}})}}{{\left( {{x^{303}} + {y^{303}}} \right)\left( {{x^{606}} + {y^{606}}} \right)}} = }

 \frac{{({x^{101}} + {y^{101}})({x^{505}} + {y^{505}})}}{{\left( {{x^{303}} + {y^{303}}} \right)\left( {{x^{606}} + {y^{606}}} \right)}} + \frac{{\left( {{y^{101}} + {z^{101}}} \right)\left( {{y^{505}} + {z^{505}}} \right)}}{{\left( {{y^{303}} + {z^{303}}} \right)\left( {{y^{606}} + {z^{606}}} \right)}} + \frac{{\left( {{z^{101}} + {x^{101}}} \right)\left( {{z^{505}} + {x^{505}}} \right)}}{{\left( {{z^{303}} + {x^{303}}} \right)\left( {{z^{606}} + {x^{606}}} \right)}}    

=3

Posted by

Nehul

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