Trig Proof

**rtblue** · April 11th 2011, 07:26 PM

Hey guys, I got stuck on this proof:

Prove that $\displaystyle \frac{secx+tanx}{cscx+cotx}=\frac{cotx-cscx}{tanx-secx}$

Any help is appreciated!

**Soroban** · April 11th 2011, 07:51 PM

Hello, rtblue!

$\text{Prove that: }\;\displaystyle \frac{\sec x+\tan x}{\csc x+\cot x}\;=\;\frac{\cot x-\csc x}{\tan x-\sec x}$

On the left side, we have: . $\displaystyle \frac{\sec x + \tan x}{\csc x + \cot x}$

Multiply by $\displaystyle \frac{(\sec x -\tan x)(\csc x - \cot x)}{(\sec x - \tan x)(\csc x - \cot x)}$

. . $\displaystyle \frac{\sec x + \tan x}{\csc x + \cot x} \cdot\frac{(\sec x - \tan x)(\csc x - \cot x)}{(\sec x - \tan x)(\csc x - \cot x)}$

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

$\displaystyle\text{Multiply by }\frac{\text{-}1}{\text{-}1}\!:\;\;\frac{\text{-}1}{\text{-}1}\cdot\frac{\csc x - \cot x}{\sec x - \tan x} \;=\;\frac{\cot x - \csc x}{\tan x - \sec x}$

**johnny** · April 11th 2011, 08:02 PM

sec x + tan x = 1/cos x + sin x / cos x = (1 + sin x)/cos x, cot x - csc x = cos x / sin x - 1/sin x = (cos x - 1)/sin x
(1 + sin x)/cos x =? (cos x - 1)/sin x
sin x + sin²x = cos²x - cos x
1 = cos x - sin x ∴ 1 + sin x = cos x, sin x = cos x - 1
∴ sec x + tan x = cot x - csc x = 1
cot x - csc x + 2 csc x = sec x + tan x + 2 csc x
2 csc x =? -2 sec x
csc x =? -sec x
ASTC for four quadrants, where sine for second quadrant and cosine for fourth quadrant
∴ csc x = -sec x
∴ 2 csc x = -2 sec x
∴ cot x - csc x + 2 csc x = sec x + tan x + 2 csc x = sec x + tan x - 2 sec x = tan x - sec x
∴ (sec x + tan x)/(csc x + cot x) = (cot x - csc x)/(tan x - sec x)

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