Question

I'm working on trig identities and in need of some help :
(1-sec x)/(1+sec x)= (cos x-1)/(cos x+1)
AND
(cos x)/(1+sin x) + (1+sin x)/ (cos x)= 2 sec x

Answer 1

Your first one can be solved by the definition of secant:\[\frac{1-\sec x}{1+ \sec x}=\frac{1-\frac{1}{\cos x}}{1+\frac{1}{\cos x}}\]

Answer 2

\[=\frac{\frac{\cos x-1}{\cos x}}{\frac{\cos x +1}{\cos x}}=\frac{\cos x -1}{\cos x +1}\]

Answer 3

(1-1/cosx)/(1+1/cosx)
=(cosx-1)/cosx/(cosx+1)/cosx
=(cosx-1)(cosx+1)

Answer 4

\[\frac{\cos x}{1+\sin x}+\frac{1+ \sin x }{\cos x}=\frac{\cos^2x+(1+\sin x)^2}{\cos x (1+\sin x)}\]

Answer 5

\[=\frac{\cos^x+1+2\sin x + \sin^2x}{\cos x(1+\sin x)}=\frac{2+2\sin x}{\cos x (1+\sin x)}\]

Answer 6

\[=\frac{2(1+\sin x)}{\cos x (1+\sin x)}\]

Answer 7

\[=\frac{2}{\cos x}=2\sec x\]

Answer 8

What are the identities used to get from this one fraction to the next?
cos x +1+2sin x+ sin ^{2} x div cos x times 1 + sin x = 2+ 2 sin x div cos x times 1 + sin x

Answer 9

I thought that's what you were writing about...\[\frac{\cos^2x+1+2\sin x+\sin^2x}{\cos x(1+\sin x)}\]\[=\frac{(\cos^2 x+ \sin^2 x)+1+2\sin x}{\cos x (1+\sin x)}\]\[=\frac{1+1+2\sin x}{\cos x(1+\sin x)}=\frac{2+2\sin x}{\cos x(1 + \sin x)}=\frac{2(1+\sin x)}{\cos x(1+\sin x)}\]

Answer 10

It's just the \[\cos^2x + \sin^2 x =1\] identity.

Answer 11

Both of you, thank you so very much for your expertise with this problems, it helps me greatly!!!! I really appreciate it! :)

Answer 12

You're welcome :) Thanks for the appreciation - not everyone bothers to say thank you!

Answer 13

I mean it, I am very greatful for the help, thanks for taking the time to help me, you deserve a thank you, you both do :)

Answer 14

my pleasure;-)
@lokisan i bother if my network doesnot fail;-)

Answer 15

my pleasure;-)
@lokisan i bother if my network doesnot fail;-)