Question

prove the above Q...........
and Explain​

Answer 1

let sin θ = x and cos θ = y

then the given equation becomes

$= > \frac{ {x}^{3} + {y}^{3} }{x + y} + \frac{ {x}^{3} - {y}^{3} }{x - y} = 2 \\ \\ taking \: lhs \: we \: have \\ \\ = \frac{ {x}^{3} + {y}^{3} }{x + y} + \frac{ {x}^{3} - {y}^{3} }{x - y} \\ \\ = \frac{(x - y)( {x}^{3} + {y}^{3}) +(x + y)( {x}^{3} - {y}^{3}) }{(x + y)(x - y)} \\ \\ = \frac{ {x}^{4} + x {y}^{3} - {x}^{3} y - {y}^{4} + {x}^{4} - x {y}^{3} + {x}^{3} y - {y}^{4} }{ {x}^{2} - {y}^{2} } \\ \\ = \frac{2( {x}^{4} - {y}^{4}) }{ {x}^{2} - {y}^{2} } \\ \\ = \frac{2( { ({x}^{2}) }^{2} - { ({y}^{2}) }^{2} )}{ {x}^{2} - {y}^{2} } \\ \\ = 2 \frac{( {x}^{2} + {y}^{2})( {x}^{2} - {y}^{2} )}{ {x}^{2} - {y}^{2} } \\ \\ = 2( {x}^{2} + {y}^{2} ) \\ \\ on \: substituting \: the \: values \: \\ \\ = 2( {sin }^{2} θ + {cos}^{2} θ) \\ \\ = 2 \times 1 \\ \\ = 2 \\ \\ = rhs$