# identity proving with two terms in denominator

• Mar 3rd 2013, 07:24 AM
ellicott
identity proving with two terms in denominator
I have a trig identity proving question:

(1 - sin(x))/(1 + sin(x)) = tan(x) + 1/cos(x)

by multiplying 1 - sin(x) on left side doesn't seems work.

Anyone can help. Thanks in advanced.
• Mar 3rd 2013, 08:02 AM
Soroban
Re: identity proving with two terms in denominator
Hello, ellicott!

Are there typos?
The statement is not true.

Quote:

$\text{Prove: }\:\frac{1 - \sin x}{1 + \sin x} \:=\: \tan x + \frac{1}{\cos x}$

$\text{Let }x = \tfrac{\pi}{6}$

$\text{The left side is: }\:\frac{1-\sin\frac{\pi}{6}}{1+\sin\frac{\pi}{6}} \:=\:\frac{1-\frac{1}{2}}{1 + \frac{1}{2}} \;=\;\frac{\frac{1}{2}}{\frac{3}{2}} \;=\;\tfrac{1}{3}$

$\text{The right side is: }\:\tan\tfrac{\pi}{6} + \sec\tfrac{\pi}{6} \;=\;\tfrac{1}{\sqrt{3}} + \tfrac{2}{\sqrt{3}} \;=\;\tfrac{3}{\sqrt{3}} \;=\;\sqrt{3}$

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I found this identity . . .

$\frac{1-\sin x}{1+\sin x} \;=\; \frac{1 -\sin x}{1+\sin x}\cdot {\color{blue}\frac{1-\sin x}{1-\sin x}} \;=\;\frac{(1-\sin x)^2}{1 - \sin^2\!x} \;=\;\frac{(1-\sin x)^2}{\cos^2\!x}$

. . . . . . $=\;\left(\frac{1-\sin x}{\cos x}\right)^2 \;=\;\left(\frac{1}{\cos x} - \frac{\sin x}{\cos x}\right)^2 \;=\;(\sec x - \tan x)^2$

Therefore: . $\frac{1-\sin x}{1+\sin x} \;=\;(\sec x - \tan x)^2$
• Mar 3rd 2013, 09:01 AM
ellicott
Re: identity proving with two terms in denominator
Thanks Soroban for correction of my typo. In general, what is the method to test true identity? Do you have to use a particular angle such as pi/6?