Math, asked by Navanihn, 1 year ago

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

Answers

Answered by Anonymous
4
\huge{\mathfrak{Answer}}

______________________________

=> \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

____________________________

\huge\sf{THANKS}

Anonymous: yep
Anonymous: kk thnq
Anonymous: yep..!!
Anonymous: oo really...!!well u thnk that some1 will give u his/her phn no. sooo easily... lol
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