Question

please please please please please answer...​

Answer 1

Answer:

I will solve ques no. 28

$given \frac{x}{a} \ \cos(x) + \frac{y}{b} \sin(x) = m \: \: \: \: = > (1)\\ \\ \: \: \: \: \: \: \: \: \: \: \: \frac{x}{a} \ \ \sin(x) + \frac{y}{b} \ \cos(x) = n \: \: \: = > (2)\\ \\ now \: squaring \: the \: first \: eq \: we \: get \: \\ \\ {m}^{2} = \frac{x^{2} }{ {a}^{2} } { \cos(x) }^{2} + \frac{y^{2} }{ {b}^{2} } { \ \sin(x) }^{2} + 2. \frac{x}{a} . \frac{y}{b} . \sin(x) \cos(x) = = \: > (3) \\ {n}^{2} = \frac{x^{2} }{ {a}^{2} } { \ \sin (x) }^{2} + \frac{y^{2} }{ {b}^{2} } { \ \ \cos(x) }^{2} - 2. \frac{x}{a} . \frac{y}{b} . \sin(x) \cos(x) = = \: > (4) \\ \\ now \: adding \: 3rd \: and \: 4th \: equations \: we \: \\ get ...... \\ {m}^{2} + {n}^{2} = \frac{x^{2} }{ {a}^{2} } { \cos(x) }^{2} + \frac{y^{2} }{ {b}^{2} } { \ \sin(x) }^{2} + 2. \frac{x}{a} . \frac{y}{b} . \sin(x) \cos(x) + \frac{x^{2} }{ {a}^{2} } { \ \sin (x) }^{2} + \frac{y^{2} }{ {b}^{2} } { \ \ \cos(x) }^{2} - \: 2. \frac{x}{a} . \frac{y}{b} . \sin(x) \cos(x) \\ {m}^{2} + {n}^{2} = \frac{x^{2} }{ {a}^{2} } \: ({ \: \: \sin( {x}^{2} ) + \cos( {x}^{2} ) }) + \frac{y^{2} }{ {b}^{2} } \: ({ \: \: \sin( {x}^{2} ) + \cos( {x}^{2} ) }) \: \\ we \: know \: that \: math \: identity \: { \: \: \sin( {x}^{2} ) + \cos( {x}^{2} ) } \: = 1\\ \: so \: \: {m}^{2} + {n}^{2} = \frac{x^{2} }{ {a}^{2} } + \frac{y^{2} }{ {b}^{2} } \: \\ hence \: proved \:$