Question

Math help!

Answer 1

yes very much needed

Answer 2

look\[tanx+secx=\frac{ sinx }{ cosx }+\frac{ 1 }{ cosx }=\frac{ sinx+1 }{ cosx }\\now~multiply~(1-sinx)~with~denominator~and ~numerator\\tanx+secx=\frac{ (1+sinx)(1-sinx) }{ cosx(1-sinx) }=\frac{ 1-\sin^2x }{ cosx(1-sinx)}\\now ~\in~the~next~steP~~we~have~\to~use\\sin^2x+\cos^2x=1\\1-\sin^2x=\cos^2x\\so\\tanx+secx=\frac{ \cos^2x }{ cosx(1-sinx) }=\frac{ cosx }{ 1-sinx }\]

Answer 3

so its true?

Answer 4

yes it should be true

Answer 5

ok thank you so much for your help! :D

Answer 6

yw!!