# Prove the following trigonometrical identity-

• Feb 1st 2010, 08:30 AM
snigdha
Prove the following trigonometrical identity-
(1- tan x + sec x)/(1+tan x + sec x) = cos x/(1+sin x)
• Feb 1st 2010, 11:19 AM
Henryt999
Hii
quick question, are you allowed to manipulate the right side of the equation? Some teachers donīt appreciate that..
What have you tried so far?
Have you tried writing $\displaystyle tan(x)= \frac{sin(x)}{cos(x)}$
and $\displaystyle sec(x) = \frac{1}{cos(x)}$
Doing so will easily give you a common denominator on the left side of the = sign...
• Feb 1st 2010, 09:37 PM
snigdha
well i'm afraid i've never tried that....if u know some other method to get this done then plz try that out...(Speechless)
• Feb 1st 2010, 10:58 PM
Soroban
Hello, snigdha!

If you know this identity: .$\displaystyle \sec^2\!x - \tan^2\!x \:=\:1$, here's another approach.

Quote:

Prove: /$\displaystyle \frac{1- \tan x + \sec x)}{1+\tan x + \sec x} \:=\: \frac{\cos x}{1+\sin x}$
We have: .$\displaystyle \frac{1 + \sec x - \tan x}{1 + \sec x +\tan x}$

Multiply by $\displaystyle \frac{\sec x + \tan x}{\sec x + \tan x}\!:\qquad \frac{\sec x + \tan x}{\sec x + \tan x} \cdot\frac{1 + \sec x - \tan x}{1 + \sec x + \tan x}$ . $\displaystyle \;\;=\;\; \frac{(\sec x + \tan x) + (\sec x + \tan x)(\sec x - \tan x)}{(\sec x + \tan x)(1 + \sec x + \tan x)}$

. . . . . $\displaystyle =\; \frac{(\sec x + \tan x) + \overbrace{(\sec^2\!x - \tan^2\!x)}^{\text{This is 1}}}{(\sec x+\tan x)(1 + \sec x +\tan x)}$ . $\displaystyle =\;\;\frac{\overbrace{\sec x + \tan x + 1}^{\text{These cancel}}}{(\sec x + \tan x)\underbrace{(1 + \sec x +\tan x)}_{\text{These cancel}}}$

. . . . . $\displaystyle =\; \frac{1}{\sec x + \tan x} \;\;=\;\;\frac{1}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}}$

Multiply by $\displaystyle \frac{\cos x}{\cos x}\!:\qquad \frac{\cos x}{\cos x}\cdot \frac{1}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}} \;=\; \frac{\cos x}{1 + \sin x}$

• Feb 1st 2010, 11:45 PM
Hello snigdha
Quote:

Originally Posted by snigdha
(1- tan x + sec x)/(1+tan x + sec x) = cos x/(1+sin x)

... and here's an approach that just uses sines and cosines.

$\displaystyle \frac{1-\tan x +\sec x}{1+\tan x + \sec x}=\frac{1-\tan x +\sec x}{1+\tan x + \sec x}\times\frac{\cos^2x}{\cos^2x}$
$\displaystyle =\frac{\cos x(\cos x-\sin x +1)}{\cos^2x+\cos x(\sin x + 1)}$

$\displaystyle =\frac{\cos x(\cos x-\sin x +1)}{1-\sin^2x+\cos x(\sin x + 1)}$

$\displaystyle =\frac{\cos x(\cos x-\sin x +1)}{(1+\sin x)(1-\sin x)+\cos x(\sin x + 1)}$

$\displaystyle =\frac{\cos x(\cos x-\sin x +1)}{(1+\sin x)(1-\sin x+\cos x)}$

$\displaystyle =\frac{\cos x}{1+\sin x}$

• Feb 2nd 2010, 10:55 PM
pacman
Soroban and Grandad did it again wih EASE, wow!

http://www1.wolframalpha.com/Calcula...image/gif&s=49

!
• Feb 3rd 2010, 12:38 AM