Question

Plz solve the above attached fig . maths ❤️❤️​

Answer 1

${ \large{ \underline{ \underline{ \rightarrow{ \mathsf{solution}}}}}}$

${ \large{ \underline{ \underline{ \rightarrow{ \mathtt{given}}}}}}$

${ \mathsf{ \rightarrow{if \: \frac{ \cos {}^{4 \: } \alpha \: }{ \cos{}^{4} \: \: \beta } + \frac{ \sin {}^{4} \alpha }{ \sin {}^{2} \beta } = 1 \: }}} \\ { \mathsf{prove \: that}}$

${ \tiny{ \mathsf{ \rightarrow{ \frac{ \cos {}^{4} \alpha }{ \cos {}^{2} \beta } = 1 - \frac{ \sin {}^{1} \alpha }{ \sin {}^{2} \beta } = \frac{ \sin {}^{2} \beta - \sin {}^{4} \alpha }{ \sin {}^{2} \beta } }}}}$

${ \tiny{ \mathsf{ \rightarrow{ \frac{ \cos {}^{4} \alpha }{ \cos {}^{2} \beta } \: = \frac{ \sin {}^{2} \beta \: = \: \sin {}^{4} \alpha }{ \sin {}^{2} \beta } }}}}$

${ \tiny{ \mathsf{ \rightarrow{ \ \cos {}^{4} \alpha \sin {}^{2} \beta = ( \sin {}^{2} \beta \sin {}^{4} \alpha ) \cos {}^{2} \beta }}}}$

${ \tiny{ \mathsf{ \rightarrow{(1 - \sin {}^{2} \alpha ) \sin {}^{2} \beta = (\sin {}^{2} \beta - \sin {}^{4} \alpha) (1 - \sin {}^{2} \beta )}}}}$

${ \tiny{ \mathsf{ \rightarrow{(1 + \sin {}^{4} \alpha = - 2 \sin {}^{2} \alpha) \sin {}^{2} \beta = \sin {}^{2} \beta - \sin {}^{4} \alpha - \sin {}^{4} \beta + \sin {}^{4} \sin {}^{2} \beta }}}}$

${ \tiny{ \mathsf{ \rightarrow{ \sin {}^{2} \beta + \sin {}^{4} \alpha \sin {}^{2} \beta - \sin {}^{2} \beta \sin {}^{2} \alpha \sin {}^{2} \beta }}}}$

${ \tiny{ \mathsf{ = \sin {}^{2} \beta - \sin {}^{4} \alpha - \sin {}^{4} \beta + \sin {}^{4} \alpha \sin {}^{2} \beta }}}$

${ \tiny{ \boxed{ \mathsf{ \rightarrow{ \sin {}^{4} \alpha + \sin {}^{4} \beta = 2 \sin {}^{2} \alpha \sin {}^{2} \beta }}}}}$

(i) (ii)

Refer Attachment

Hope it helps u!!