Help me prove this Trigonometry identity

[footprints]

Someone pls help me prove this identity. I'm going nuts
$$tan^2x - sin^2x = tan^2x sin^2x$$

 

[Muzza]

What have you tried so far? For example, what if you rewrite everything in terms of sine and cosine (often a good thing to do):

sin^2(x) / cos^2(x) - sin^2(x) = sin^2(x)/cos^2(x) * sin^2(x)
<=>
sin^2(x) / cos^2(x) - sin^2(x)cos^2(x)/cos^2(x) = sin^4(x)/cos^2(x)
<=>
(sin^2(x) - sin^2(x)cos^2(x)) / cos^2(x) = sin^4(x)/cos^2(x)

Any ideas now? Factoring, and the Pythagorean identity will come in handy.

*edit* And after doing this, a better proof (where you only have to work with one side) appeared to me. This is often the case with trigonometric identities - at first you have a horrible mess of trigonometric expressions and after expanding, you see a way to transform one side into the other. I won't spoil the surprise though ;)

 

[footprints]

Thats what i tried/usually do but i just can't get it for this one. One question, I don't get one part,
sin^2(x) / cos^2(x) - sin^2(x)cos^2(x)/cos^2(x) = sin^4(x)/cos^2(x)
and so on.

 

[Muzza]

I'm just writing it as one fraction.

a/b - c = a/b - cb/b = (a - cb)/b, for all a, b, c.

 

[footprints]

I see. But how does it help?

 

[Muzza]

(sin^2(x) - sin^2(x)cos^2(x)) / cos^2(x) =
sin^2(x)(1 - cos^2(x)) / cos^2(x).

Recognize anything?

 

[footprints]

Oh right
So I'll get $$\frac{sin^2x(1-cos^2x)}{cos^2x} = \frac{sin^2x(1-cos^2x)}{cos^2x}$$Right?
So its proved.

 

[JonF]

I just worked it out, this one seems to be easier if you start with tan^2(x)*sin^2(x) instead of tan^2(x) – sin^2(x)

 

[footprints]

So if i start with $${tan^2x} \cdot {sin^2x}$$ would do i do from here that would make it easier?

 

[robphy]

Factor out $\tan^2 x$ on the left-hand-side.

$$\tan^2 x (1-\cos^2 x)=\tan^2 x\sin^2 x$$

 

[jai6638]

Oh right
So I'll get $$\frac{sin^2x(1-cos^2x)}{cos^2x} = \frac{sin^2x(1-cos^2x)}{cos^2x}$$Right?
So its proved.

but when you write $$\frac{sin^2x(1-cos^2x)}{cos^2x} = \frac{sin^2x(1-cos^2x)}{cos^2x}$$

you get Sin^2/Cos^2 x ( 1-cos^2)/Cos^2 so how does the second part form the sin^2x?? it would again land up forming Tan^2x ( because its sin^2/cos^2 )

thanks

 

[footprints]

I don't understand what u mean. When u say second part, do u mean $$\frac{1 - cos^2x}{cos^2x}$$? Or the right hand-side?

 

[ChanDdoi]

i think that is what jai6638 meant, footprints

jai6638, you forgot that A*B / C = A/C * B and does not equal to A/C * B/C

 

[JonF]

if you can get to:
$$\frac{sin^2x(1-cos^2x)}{cos^2x}$$


Distribute the sin in, and then break the fraction apart over addition

 

[nolachrymose]

There's no reason to distribute, though. The basic identity that $\sin^2x=1-cos^2x$ can be applied to get:

$$\frac{\sin^2x*\sin^2x}{cos^2x} &= \tan^2x\sin^2x$$

 

[jai6638]

i think that is what jai6638 meant, footprints

jai6638, you forgot that A*B / C = A/C * B and does not equal to A/C * B/C

lol... yeah that's what i was thinkin for some odd reason.. damn! thanks a lot ... need to start doing my trignometry again.. doing algebra II at my new school... sucks ass... my previous school was soo much better .. we did our trignometry in 9th grade ..

cheers