Question

prove that the question in attachment​

Answer 1

solution of your question is In the above image.

Answer 2

$\huge\star{\underline{\mathfrak{Answer}}}$

$\alpha + \beta = 90 {}^{o}$

$\beta = 90 {}^{o} - \alpha$

LHS=

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

$= \sqrt{ \cos( \alpha ) \csc( {90}^{o} - \beta ) - \cos( \alpha ) \sin( {90}^{o} - \beta ) }$

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

$= \sqrt{1 - \cos {}^{2} ( \alpha ) }$

$= \sqrt{ \sin {}^{2} ( \alpha ) }$

$= \sin( \alpha ) = rhs$

FORMULAS USED:

• $\csc(90 {}^{o} - a ) = \sec(a)$
• $= \sin(90 {}^{o} - a ) = \cos(a)$
• $cos(a) \times \sec(a) = 1$
• $\sin {}^{2} (a) + \cos {}^{2} (a) = 1$