Question

○●● No. 61 please ●●○

Answer 1

SOLUTION:-

Given:

${x}^{m} = {a}^{m} \: {cos}^{4} \theta \: \: and \: \: {y}^{n} = {b}^{m} \: {sin}^{4} \theta$

Therefore,

$\frac{ {x}^{n} }{ {a}^{m} } = {cos}^{4} \theta$

We can write as,

${cos}^{2} \theta = ( \frac{ {x}^{n} }{ {a}^{m} } ) {}^{ \frac{1}{2} }$

Similarly,

$= > \frac{ {y}^{n} }{ {b}^{m} } = {sin}^{4} \theta \\ \\ = > {sin}^{2} \theta =( \frac{ {y}^{n} }{ {b}^{m} } ) {}^{ \frac{1}{2} }$

We know that;

${sin}^{2} \theta + {cos}^{2} \theta = 1 \\ \\ so \\ = > (\frac{ {y}^{ \frac{n}{2} } }{ {b}^{ \frac{m}{2} } } + \frac{ {x}^{ \frac{n}{2} } }{ {a}^{ \frac{m}{2} } } ) = 1$

Hence,

Proved.

Option (a)✓

Hope it helps ☺️