Question

verify the identity (cosx/1+sinx)+(1+sinx/cosx)= 2 secx

Answer 1

=(1+ sinx)/(cos x) + [(cos x)(1-sinx)]/[(1+sin x)(1-sinx)]

=(1+ sinx)/(cos x) + [(cos x)(1-sinx)]/(1-sin²x)

=(1+ sinx)/(cos x) + [(cos x)(1-sinx)]/(cos²x)

=(1+ sinx)/(cos x) + (1-sinx)/(cosx)

=1/cosx + tanx + 1/cosx - tanx

=secx +secx

=2secx

Answer 2

is this correct?

Answer 3

wondering where [(cos x)(1-sinx)]/[(1+sin x)(1-sinx)] came from

Answer 4

\(\bf \cfrac{cos(x)}{1+sin(x)}+\cfrac{1+sin(x)}{cos(x)}=2sec(x)\) right?

Answer 5

yep

Answer 6

one sec

Answer 7

\(\bf \cfrac{cos(x)}{1+sin(x)}+\cfrac{1+sin(x)}{cos(x)}\implies \cfrac{cos^2(x)+[1+sin(x)]^2}{[1+sin(x)]cos(x)}
\\ \quad \\
\cfrac{cos^2(x)+{\color{brown}{ 1^2+2sin(x)+sin^2(x) }}}{[1+sin(x)]cos(x)}
\\ \quad \\
\cfrac{cos^2(x)+sin^2(x)+1+2sin(x) }{[1+sin(x)]cos(x)}
\\ \quad \\
\cfrac{{\color{brown}{ 1}}+1+2sin(x) }{[1+sin(x)]cos(x)}\implies \cfrac{2+2sin(x) }{[1+sin(x)]cos(x)}
\\ \quad \\
\cfrac{2\cancel{[1+sin(x)]} }{\cancel{[1+sin(x)]} cos(x)}\impliedby \textit{taking common factor atop}\)

and surely you can see what that is

Answer 8

then you get 2secx thanks :)