Question

(1+cos x+sin x)/ (1+cos x-sin x)= Sec x+Tan x

Answer 1

Right side = 1/cosx + sinx / cos x = ( 1 + sinx) / cos x)

i'll write sin x as s and cos x as c

so left side = (1 + c+ s) / (1 + c- s) = (1 + s) / c (right side)

Answer 2

I got that for the right side but I can't figure out how to get the left to that point as well

Answer 3

now if we cross multiply
c(1 + c+s) = (1 + s)(1 + c- s)
c + c^2 + cs = 1 + c - s + s + sc - s^2
c + cs = 1 + c - s + s + sc - (s^2 + c^2)

s^2 + c^2 = sin^2 x + cos^2 x = 1

so c + cs = 1 -1 - s+ s + c + cs
c + cs = c + cs


which completes the proof

Answer 4

you need to simplify and use the identity sin^x = cos^x = 1

Answer 5

\(\large\tt \begin{align} \color{black}{\dfrac{1+cosx+sinx}{1+cosx-sinx}\\~\\
=\dfrac{1+cosx+sinx}{1+cosx-sinx}\times \dfrac{secx}{secx}\\~\\
=\dfrac{secx+tanx+1}{secx-tanx+1}\\~\\
=\dfrac{secx+tanx+(sec^2x-tan^2x)}{secx-tanx+1}\\~\\
=\dfrac{(secx+tanx)(1+secx-tanx)}{secx-tanx+1}\\~\\
=(secx+tanx)\\~\\}\end{align}\)

Answer 6

mathmath333 why would you multiply by secx/secx?

Answer 7

because that will give the next step
\(\Large =\dfrac{secx+tanx+1}{secx-tanx+1}\\~\\\)

which will be easy for ur proof

Answer 8

\(\large \begin{align} \color{black}{remember\\
\Large cosx\times secx=1\\~\\
\Large sinx\times secx=tanx}\end{align}\)