Prove the following trigonometrical identity-

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February 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)
February 1st 2010, 11:19 AM
Henryt999

Hii

quick question, are you allowed to manipulate the right side of the equation? Some teachers don´t appreciate that..
What have you tried so far?
Have you tried writing $tan(x)= \frac{sin(x)}{cos(x)}$
and $sec(x) = \frac{1}{cos(x)}$
Doing so will easily give you a common denominator on the left side of the = sign...
February 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)
February 1st 2010, 10:58 PM
Soroban

Hello, snigdha!

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

Quote:

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

We have: . $\frac{1 + \sec x - \tan x}{1 + \sec x +\tan x}$

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

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

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

Multiply by $\frac{\cos x}{\cos x}\!:\qquad \frac{\cos x}{\cos x}\cdot \frac{1}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}} \;=\; \frac{\cos x}{1 + \sin x}$
February 1st 2010, 11:45 PM
Grandad

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.

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

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

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

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

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

Grandad
February 2nd 2010, 10:55 PM
pacman

Soroban and Grandad did it again wih EASE, wow!

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

!
February 3rd 2010, 12:38 AM
Grandad

Quote:

Originally Posted by pacman

Soroban and Grandad did it again wih EASE, wow!

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

!

... That's what comes of a lifetime of useless activity!

Grandad