Question

Any one can solve this questuon

Answer 1

Interesting question =_+

${ \sin }^{2} \alpha = \dfrac{{x}^{2} + {y}^{2} }{2xy} \\ \\0 \leqslant { \sin }^{2} \alpha \leqslant 1 \\ 0 \leqslant \dfrac{{x}^{2} + {y}^{2} }{2xy} \leqslant 1 \\ \\ 0 \leqslant {x}^{2} + {y}^{2} \leqslant 2xy \\ \\ {x}^{2} + {y}^{2} - 2xy \leqslant 0 \\ \\ {(x - y)}^{2} \leqslant 0$
Since square of a number can NOT be negative ,

$( {x - y)}^{2} = 0 \\ \\ x = y$
Proved.

Also , x = y ≠ 0 elsewise , it will be undefined.