Question

prove that, cos theta ÷ 1-sin theta + cos theta ÷1+sin theta = 2 sec theta​

Answer 1

${ \red{ \bold{ \huge{ \underline{Question:-}}}}}$



▪ Prove that ,



${ \bold{ \frac{ \cos(theta) }{1 - \sin(theta) } + \frac{ \cos(theta) }{1 - \sin(theta) } = 2 \sec(theta) }}$



${ \red{ \huge{ \bold{ \underline{Solution:-}}}}}$




▪ firstly, let theta be equal to  α just for my convenience....


thus , we have to proof...



${ \bold{ \purple{ \frac{ \cos( \alpha ) }{1 - \sin( \alpha ) } + \frac{ \cos( \alpha ) }{1 + \sin( \alpha ) } = 2 \sec( \alpha ) }}}$



▪ solving the L.H.S......



${ \bold{ = \frac{ \cos( \alpha ) }{1 - \sin( \alpha ) } + \frac{ \cos( \alpha ) }{1 + \sin( \alpha ) } }}$




▪ Taking LCM ....



${ \bold{ = \frac{ \cos( \alpha )(1 + \sin( \alpha ) ) + \cos( \alpha ) (1 - \sin( \alpha )) }{(1 - \sin( \alpha ) )(1 + \sin( \alpha )) } }}$




${ \bold{ = \frac{ \cos( \alpha ) + \cos( \alpha )\sin( \alpha ) + \cos( \alpha) - \cos( \alpha) \sin( \alpha )}{(1 - \sin( \alpha ) )(1 + \sin( \alpha ))}}}$



${ \bold{ = \frac{2 \cos( \alpha ) }{(1 - \sin( \alpha) )(1 + \sin( \alpha )) } }}$




▪ using the algebraic identity in the denominator.....




${ \boxed{ \bold{ \red{ \: (a - b)(a + b) = {a}^{2} - {b}^{2} \: }}}}$



${ \bold{ = \frac{2 \cos( \alpha ) }{1 - { \sin( \alpha ) }^{2} } }}$




▪ we know that.....




${ \boxed{ \bold{ \red{ \: 1 - { \sin( \alpha ) }^{2} = { \cos( \alpha ) }^{2} \: }}}}$




▪ thus ,substituting this value....we get-



${ \bold{ = \frac{2 \cos( \alpha ) }{ { \cos( \alpha ) }^{2} } }}$




${ \bold{ = \frac{2}{ \cos( \alpha ) } }}$



▪ using the trigonometric identity....




${ \boxed{ \bold { \red{ \: \: \frac{1}{ \cos( \alpha ) } = \sec( \alpha ) \: }}}}$




${ \bold{ = 2 \sec( \alpha ) }}$



thus,



☆ L.H.S = R.H.S.



hence, proved that



${ \bold{ \red{ \frac{ \cos(theta) }{1 - \sin(theta) } + \frac{ \cos(theta) }{1 + \sin(theta) } = 2 \sec(theta) }}}$