Solving an Equation with Trigonometric Terms

[sacwchiri]

more trig...

Amm I've been at this problem for an hour already it loooked really easy but for some reason i can't reach the answer
(tan x + sec x)^2 = (1 + sin x)/(1 - sin x)

soo what went and tried was expand it and then exchage tan and sec..

(sin^2 x /cos^2 x )+ (2/cos x) + (1/ sin^2 x)

((sin^2)(sin^2) + (2(sin^2)(cos)) + cos^2)/((sin^2)(cos^2))

((sin^2)(sin^2 + 2cos - 1) + 1)/((sin^2)(1-sin^2))

(sin^2 + 2cos)/(1-sin^2)

and after i get there i njust don't know how to get to the (1 + sin x)/(1 - sin x)

any ideas??

 

[malawi_glenn]

you can write

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

 

[VietDao29]

Amm I've been at this problem for an hour already it loooked really easy but for some reason i can't reach the answer
(tan x + sec x)^2 = (1 + sin x)/(1 - sin x)

soo what went and tried was expand it and then exchage tan and sec..

(sin^2 x /cos^2 x )+ (2/cos x) + (1/ sin^2 x)

Oh, you're wrong from the start... =.="
sec(x) = 1 / (cos(x)), instead of 1 / (sin(x)), as you have written above.
You should re-do the problem, and follow malawi_glenn's hint.

You can also manipulate both sides a the same time, instead of 1 side at 1 time. Like this:
(\tan x + \sec x) ^ 2 = \frac{1 + \sin x}{1 - \sin x}
\Leftrightarrow \left (\tan x + \frac{1}{\cos x} \right) ^ 2 = \frac{(1 + \sin x) ^ 2}{1 - \sin ^ 2 x}
<=> ...

Can you go from here? :) It should be easy.

 

[sacwchiri]

ok well after asking arround i got to the answer... thanks for the help but the thing is I am supposed to just use the properties to reach other side... i can't alter it... but i guess is my fault for not specifying... well the way it went was quite tricky and required a property i don't use normally which is tan^2x + 1 = sec^2 x

so using that you get that

2tan x sec x + sec^2x - 1... and one more thing i realized i was using the sec = 1/sin ... ummm opps... but really thanks for the ideas