Question

(sec ø+ tan ø) (1-sin ø)=cos ø​

Answer 1

$\huge \mathfrak{ \red{solution}}$

LHS :

$\star \:( \sec( \alpha ) + \tan( \alpha ) )(1 - \sin( \alpha ) ) \\ \\ \star \: ( \frac{1}{ \cos( \alpha ) } + \frac{ \sin( \alpha ) }{ \cos( \alpha ) } )(1 - \sin( \alpha ) ) \\ \\ \star \: ( \frac{1 + \sin( \alpha ) }{ { \cos}\alpha }) (1 - \sin( \alpha ) ) \\ \\ \star \: \frac{ {1}^{2} - { \sin}^{2} \alpha }{ { \cos} \alpha } \\ \\ \star \: \frac{ { \cos}^{2} \alpha }{ \cos( \alpha ) } \\ \\ \star \: \cos( \alpha )$

LHS = RHS

$\star \: \huge \boxed{ \blue{proved}}$

Answer 2

$( \sec ø+ \tan ø) (1- \sin ø)= \cos ø$

$\tt LHS :$

$\implies ( \frac{1}{ \cos \phi } + \frac{ \sin \phi}{ \cos\phi } )(1 - \sin \phi)$

$\implies ( \frac{1 + \sin \phi }{ \cos \phi } )(1 - \sin \phi)$

$\implies \frac{1 - \sin^{2}\phi }{ \cos \phi }$

$\tt we \: know \: that :$

$\tt\red{ \boxed{ 1 - \sin^{2} \theta = \cos^{2} \theta }}$

$\implies \frac{ { \cos }^{ \bcancel2} \phi }{ \bcancel{\cos \phi}}$

$\implies \cos \phi$

$\orange{ \therefore LHS = RHS}$

$\purple{\therefore( \sec ø+ \tan ø) (1- \sin ø)= \cos ø}$

$\red{\underline{\underline{Hence\:Proved}}}$