Hey does anyone know these trig idenities??
((Cosx +1)/(secx-tanx)) - ((1-cosx)/secx+tanx)) = 2(1+tanx)
sec(-x) / sin(180-x) = tan(270-x) + tan(180+x)
(tanx-sinx) / (sin(cubed)x) = (sec) / (1+cosx)
Hello Giggly2If you mean do I know how to prove them, yes.
For this one, you need to know that if you divide both sides of $\displaystyle \cos^2x +\sin^2x = 1$ by $\displaystyle \cos^2x$, you get:((Cosx +1)/(secx-tanx)) - ((1-cosx)/secx+tanx)) = 2(1+tanx)
$\displaystyle 1+\tan^2x = \sec^2x $
$\displaystyle \Rightarrow \sec^2x-\tan^2x = 1$
$\displaystyle \Rightarrow (\sec x+\tan x)(\sec x - \tan x) = 1$
Then just put both the fractions on the LHS over a common denominator, multiply out the brackets (be careful with the minus signs!) and you'll get $\displaystyle 2(1 + \tan x)$. I'll start you off:
$\displaystyle \frac{\cos x +1}{\sec x -\tan x}-\frac{\cos x -1}{\sec x +\tan x} = \frac{(\cos x +1)(\sec x +\tan x)-(1-\cos x)(\sec x - \tan x)}{(\sec x+\tan x)(\sec x - \tan x)}$
$\displaystyle = \frac{\dots \quad \dots \quad \dots \quad \dots}{1}$
Can you complete it now?
You need to know the following identities:sec(-x) / sin(180-x) = tan(270-x) + tan(180+x)
- $\displaystyle \sec(-x) = \sec x$
- $\displaystyle \sin (180 - x) = \sin x$
- $\displaystyle \tan(180 + x) = \tan x$
- So $\displaystyle \tan(270 - x) = \tan(180 + 90 - x)=\tan(90-x) = \cot x$
Then $\displaystyle \frac{\sec(-x)}{\sin(180-x)}= \frac{\sec{x}}{\sin x} = \frac{1}{\cos x\sin x}$
Now
- start from the RHS using the last two identities above
- express $\displaystyle \cot x$ and $\displaystyle \tan x$ in terms of sine and cosine
- add the two fractions
- use $\displaystyle \cos^2x +\sin^2x = 1$
and you're there.
Again, I'll start you off:(tanx-sinx) / (sin(cubed)x) = (sec) / (1+cosx)
$\displaystyle \frac{\tan x - \sin x}{\sin^3x}= \frac{\frac{\sin x}{\cos x}-\sin x}{\sin^3x}$
$\displaystyle = \frac{\sin x - \sin x \cos x}{\cos x\sin^3x}$
Now you need to:
- factorise the numerator
- divide top and bottom by the common factor $\displaystyle \sin x$
- use $\displaystyle \sin^2x = 1-\cos^2x =(1+\cos x)(1-\cos x)$
- divided top and bottom by a common factor again
You should end up with $\displaystyle \frac{1}{\cos x(1+\cos x)}= \frac{\sec x}{1+\cos x}$
Can you finish them off now?
Grandad