Trigo Identities....

**cannon** · September 3rd 2009, 02:21 AM

2tanx/1-tan^2x = 2/cotx-tanx

csc^2x/cox^2x = sec^2x + csc^2

(sinx+cosx)(tanx+cotx) = secx+cscx

and

1-sinx/1+sinx = (secx - tanx)^2

Thanks alot

**sa-ri-ga-ma** · September 3rd 2009, 03:31 AM

2tanx/1-tan^2x = 2/cotx-tanx
2tanx/1-tan^2x = tan2x = sin2x/cos2x
= 2sinx*cosx/ (cos^2x - sin^2x). Divide numerator and denomenator by sinx*cosx, you get
= 2/(cotx - tanx)

csc^2x/cox^2x = sec^2x + csc^2
sec^2x + csc^2
= 1/cos^2x + 1/sin^2x
= (sin^2x + cos^2x)/ sin^2x*cos^2x
= 1/(sin^2x*cos^2x)
= csc^2x/cos^2x

**cannon** · September 3rd 2009, 04:09 AM

thanks, how about the 3rd problem?

**cannon** · September 3rd 2009, 04:14 AM

Originally Posted by sa-ri-ga-ma

2tanx/1-tan^2x = 2/cotx-tanx
2tanx/1-tan^2x = tan2x = sin2x/cos2x
= 2sinx*cosx/ (cos^2x - sin^2x). Divide numerator and denomenator

what happened in the 2sinx*cosx? why did it became like did? i thought tan= sin/cos and not sin*cos? please clarify

**Soroban** · September 3rd 2009, 05:05 AM

Hello, cannon!

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

Multiply the left side by: . $\frac{1-\sin x}{1-\sin x}$

. . $\frac{1-\sin x}{1 + \sin x}\cdot{\color{blue}\frac{1-\sin x}{1-\sin x}} \;\;=\;\;\frac{(1-\sin x)^2}{1-\sin^2\!x} \;\;=\;\; \frac{(1-\sin x)^2}{\cos^2\!x}$ . $= \;\left(\frac{1-\sin x}{\cos x}\right)^2 \;=\;\left(\frac{1}{\cos x} - \frac{\sin x}{\cos x}\right)^2 \;=\;(\sec x - \tan x)^2$

**Grandad** · September 3rd 2009, 05:21 AM

Hello cannon

Originally Posted by cannon

2tanx/1-tan^2x = 2/cotx-tanx

Originally Posted by cannon

what happened in the 2sinx*cosx? why did it became like did? i thought tan= sin/cos and not sin*cos? please clarify

$\frac{2\sin x\cos x}{\cos^2x-\sin^2x}=\frac{\dfrac{2\sin x\cos x}{\color{red}\sin x \cos x}}{\dfrac{\cos^2x-\sin^2x}{\color{red}\sin x\cos x}}$

$=\frac{2}{\dfrac{\cos^2x}{\sin x\cos x}-\dfrac{\sin^2x}{\sin x\cos x}}=\frac{2}{\dfrac{\cos x}{\sin x}-\dfrac{\sin x}{\cos x}}=\frac{2}{\cot x - \tan x}$

Originally Posted by cannon

thanks, how about the 3rd problem?

$(\sin x +\cos x)(\tan x + \cot x) = (\sin x +\cos x)\Big(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\Big)$

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

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

$=\sec x + \csc x$

Grandad

**cannon** · September 3rd 2009, 05:55 PM

how did you get cos^2x-sin^2x?

**cannon** · September 3rd 2009, 06:13 PM

and for the last 2 problems hehe...

cosx/1+sinx = 1-sinx/cosx

sinx/1-cotx + cosx/1-tanxx = sinx +cosx

**Grandad** · September 3rd 2009, 10:05 PM

Hello cannon

Originally Posted by cannon

how did you get cos^2x-sin^2x?

I was picking up on the reply by sa-ri-ga-ma:

Originally Posted by sa-ri-ga-ma

2tanx/1-tan^2x = 2/cotx-tanx
2tanx/1-tan^2x = tan2x = sin2x/cos2x
= 2sinx*cosx/ (cos^2x - sin^2x). Divide numerator and denomenator by sinx*cosx, you get
= 2/(cotx - tanx)

and your query

Originally Posted by cannon

what happened in the 2sinx*cosx? why did it became like did? i thought tan= sin/cos and not sin*cos? please clarify

As far as your next post is concerned:

Originally Posted by cannon

and for the last 2 problems hehe...

cosx/1+sinx = 1-sinx/cosx

sinx/1-cotx + cosx/1-tanxx = sinx +cosx

this thread is now too complicated! Please have another attempt at these questions - they are very similar to the ones we've shown you already. If you still can't do them, re-post them and show us your attempts in a new thread. (See Rule #14).

Grandad

**cannon** · September 4th 2009, 03:57 AM

I didn't get what you mean in your first 2 statements. Anyways, I already solved the last problem, sorry for being lazy lol and thanks a lot!

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