Question

solve it plzz no spam​

Answer 1

Step-by-step explanation:

$given \: \alpha + \beta = 90 \\ taking \: cos \: of \: both \: sides \\ \cos( \alpha + \beta ) = \cos(90) \\ \cos( \alpha ) \cos( \beta ) - \sin( \alpha ) \sin( \beta ) = 0 \\ \cos( \alpha ) \cos( \beta ) = \sin( \alpha ) \sin( \beta ) \\ \cos( \alpha ) = \frac{ \sin( \alpha ) \sin( \beta ) }{ \cos( \beta ) } \\ on \: putting \: this \: value \: in \: lhs \\ \sqrt{ \frac{ \sin( \alpha ) \sin( \beta ) }{ \cos( \beta ) } \times \csc( \beta ) - \frac{ \sin( \alpha ) \sin( \beta ) }{ \cos( \beta ) } \times \sin( \beta ) } \\ \sqrt{ \frac{ \sin( \alpha ) \sin( \beta ) }{ \cos( \beta ) } \times \frac{1}{ \sin( \beta ) } - \frac{ \sin( \alpha ) { \sin( \beta ) }^{2} }{ \cos( \beta ) } } \\ \sqrt{ \frac{ \sin( \beta ) }{ \cos( \beta ) } - \frac{ \sin( \alpha ) { \sin( \beta ) }^{2} }{ \cos( \beta ) } } \\ \sqrt{ \frac{ \sin( \beta ) - \sin( \alpha ) { \sin( \beta ) }^{2} }{ \cos( \beta ) } } \\ \sqrt{ \frac{ \sin( \beta ) (}{?} }$

Answer 2

$givn$

$\alpha + \beta = 90$

$to \: prove$

$\sqrt{ \cos( \alpha ) \ \csc( \beta ) - \cos( \alpha ) \ \sin ( \beta ) } = \sin( \alpha )$

$from \: above \:$

$\alpha = 90 - \beta$

$and$

$\beta = 90 - \alpha$

$lhs$

$\sqrt{ \cos( \alpha ) \ \csc ( \beta ) - \cos( \alpha ) \sin( \beta ) }$

$putting \: the \: above \: value \: \alpha = 90 - \beta$

$\sqrt{ \cos( \alpha ) \csc(( 90 - \alpha ) - \cos( \alpha ) \sin (90 - \alpha ) }$

$\sqrt{ \cos( \alpha ) \sec( \beta ) - \cos( \alpha ) \cos( \alpha ) }$

$\sqrt{1 - \cos ^{2} \alpha }$

$\sqrt{ \sin^{2 } \alpha }$

$\sin( \alpha )$

$hence \: lhs = rhs$

proved