Question

secA(1-sinA) (secA+tanA)=1

Answer 1

$\huge{\mathfrak{Answer}}$

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=>$\sec(a) (1 - \sin(a)) ( \sec(a) + \tan(a) ) = 1$

$\underline\bold{We\:know}$

=>$\sec(a) = \frac{1}{ \cos(a) } \\ \tan(a) = \frac{ \sin(a) }{ \cos(a) }$

=>$\frac{1}{ \cos(a) } (1 - \sin(a) )( \frac{1}{ \cos(a) } + \frac{ \sin(a) }{ \cos(a) } )$

=>$\frac{1}{ \cos(a) } (1 - \sin(a) )( \frac{1 + \sin(a) }{ \cos(a) } )$

=>$\frac{1}{ \cos(a) } (1 - \sin(a))( \frac{1}{ \cos(a) } )(1 + \sin(a) )$

=>$\frac{1}{ { \cos(a) }^{2} } (1 - \sin(a)(1 + \sin(a) )$

=>$\underline\bold{By\:usng\:this\:formulae}$

=>$( {1}^{2} - { \sin(a) }^{2} ) = (1 - \sin(a) )(1 + \sin(a) )$

=>$\frac{1}{ { \cos(a) }^{2} } (1 - { \sin(a) }^{2} )$

=>$\underline\bold{We\:know}$
$(1 - { \sin(a) }^{2} ) = { \cos(a) }^{2}$

=>$\frac{1}{ { \cos(a) }^{2} } ( { \cos(a) }^{2} )$

=>$1$

$L.H.S\:=\:R. H. S$

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$\huge\sf{THANKS}$