verifying idenitity

**wolfxpax** · May 6th 2009, 07:51 PM

how would i verify this identity?
(1+sec(-x))/(sin(-X)+tan(-x))= -csc x

**Grandad** · May 6th 2009, 10:25 PM

Hello wolfxpax

Originally Posted by wolfxpax

how would i verify this identity?
(1+sec(-x))/(sin(-X)+tan(-x))= -csc x

First you need to know how to handle the negative signs:

$\sin(-x) = -\sin x$
$\cos(-x) = \cos x$

So it follows that $\tan(-x) =-\tan x, \sec(-x) = \sec x$ , and so on.

So $\frac{1 +\sec(-x)}{\sin(-x)+\tan(-x)} =\frac{1 +\sec(x)}{-\sin(x)-\tan(x)}$

$= \frac{1+\sec x}{-\sin x - \dfrac{\sin x}{\cos x}}$

$= \frac{1+\sec x}{-\sin x - \dfrac{\sin x}{\cos x}}\times\color{red}\frac{\cos x}{\cos x}$

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

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

$= \frac{1}{-\sin x}$

$=-\csc x$

Grandad

**Soroban** · May 7th 2009, 02:06 PM

Hello, wolfxpax!

A slightly different approach . . .

$\frac{1+\sec(\text{-}x)}{\sin(\text{-}x)+\tan(\text{-}x)} \:=\: -\csc x$

As Grandad pointed out: . $\begin{array}{ccc} \sec(\text{-}x) &=& \sec(x) \\ \sin(\text{-}x) &=& \text{-}\sin(x) \\ \tan(\text{-}x) &=& \text{-}\tan(x) \end{array}$

So we have: . $\frac{1+\sec x}{\text{-}\sin x - \tan x} \;=\;\frac{1+\sec x}{\text{-}\sin x - \frac{\sin x}{\cos x}} \;=\;\frac{1+\sec x}{\text{-}\sin x\left(1 + \frac{1}{\cos x}\right)}$

. . . . . . . $= \;\frac{1 + \sec x}{\text{-}\sin x(1 + \sec x)} \;=\;\frac{1}{-\sin x} \;=\;-\csc x$

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