need help with proving this trig identity

[abel muroi]

I was given this tan² x - sin² x = tan² x sin² x and i was told to prove it.



here is what i have so far... (( i started with the left side first))

tan² x - sin² x = tan² x sin² x
(sin² x)/(cos² x) - sin² x
(sin² x - sin² x)/(cos² x)


so this is as far as i can get, what should i do next?

 

[Steven G]

I was given this tan² x - sin² x = tan² x sin² x and i was told to prove it.



here is what i have so far... (( i started with the left side first))

1tan² x - sin² x = tan² x sin² x
2(sin² x)/(cos² x) - sin² x
3(sin² x - sin² x)/(cos² x)


so this is as far as i can get, what should i do next?

1st of all you should know that (sin² x - sin² x)=0. Come on.
I numbered the lines above in red. In 2, the left (sin² x) is being divided by (cos² x) while the right (sin² x) is NOT. So 3 is nonsense.

Is 5/3 - 5 = (5-5)/3 =0. Unless 5/3 = 5 when you subtract them you will not get 0.
a/b - a = a/ b - ab/b =(a-ab)/b = a(1-b)/b or another way to think about it-- a/b-a= a/b - 1a = a(1/b - 1) = a(1-b)/b

 

[abel muroi]

1st of all you should know that (sin² x - sin² x)=0. Come on.
I numbered the lines above in red. In 2, the left (sin² x) is being divided by (cos² x) while the right (sin² x) is NOT. So 3 is nonsense.

Is 5/3 - 5 = (5-5)/3 =0. Unless 5/3 = 5 when you subtract them you will not get 0.
a/b - a = a/ b - ab/b =(a-ab)/b = a(1-b)/b or another way to think about it-- a/b-a= a/b - 1a = a(1/b - 1) = a(1-b)/b

ok i re-did the problem..


tan² x - sin² x = tan² x sin² x
(sin² x)/(cos² x) - sin² x
(sin² x - sin² x cos² x)/(cos² x)
(sin² x (-1 + cos² x)) /(cos² x)


so what now?

 

[Steven G]

ok i re-did the problem..


tan² x - sin² x = tan² x sin² x
(sin² x)/(cos² x) - sin² x
(sin² x - sin² x cos² x)/(cos² x)
(sin² x (-1 + cos² x)) /(cos² x)


so what now?

Line 4 does not follow from line 3.
You should now what -1 + cos² x equals

 

[abel muroi]

Line 4 does not follow from line 3.
You should now what -1 + cos² x equals

tan² x - sin² x = tan² x sin² x
(sin² x)/(cos² x) - sin² x
(sin² x - sin² x cos² x)/(cos² x)
(sin² x + sin² x (-1 + cos² x)) /(cos² x)
(sin² x + sin² x ( sin² x))/(cos² x)


am i doing this correctly so far?

 

[Steven G]

tan² x - sin² x = tan² x sin² x
1(sin² x)/(cos² x) - sin² x
2(sin² x - sin² x cos² x)/(cos² x)
3(sin² x + sin² x (-1 + cos² x)) /(cos² x)
4(sin² x + sin² x ( sin² x))/(cos² x)


am i doing this correctly so far?

Nope, it is still wrong. Abel, you really can't study precalculus and trigonometry before you master algebra. I know you study hard, but you need to study algebra and after you finish studying algebra you study some more algebra.

From line 3 to line 4 you replaced (-1 + cos² x) with sin² x. Are they really equal?
From 2 to 3 you still have not factored correctly. In fact, line 3 (the way you have it) is simply sin² x. Do you see that?

 

[abel muroi]

Nope, it is still wrong. Abel, you really can't study precalculus and trigonometry before you master algebra. I know you study hard, but you need to study algebra and after you finish studying algebra you study some more algebra.

From line 3 to line 4 you replaced (-1 + cos² x) with sin² x. Are they really equal?
From 2 to 3 you still have not factored correctly. In fact, line 3 (the way you have it) is simply sin² x. Do you see that?

Yes i am considering retaking intermediate algebra before i start attending a csu


this time i think i have factored correctly..

tan² x - sin² x = tan² x sin² x
1(sin² x)/(cos² x) - sin² x
2(sin² x)/(cos² x) - (sin² x cos² x)/(cos² x)
3(sin² x - sin² x cos² x)/(cos² x)
4(sin² x (1 - cos² x))/(cos² x)
5(sin² x (sin² x))/(cos² x)
6(sin² x sin² x)/ (cos² x)


ok after line 6 i am a little stuck, should i break up the fraction into two?

 

[Steven G]

Yes i am considering retaking intermediate algebra before i start attending a csu


this time i think i have factored correctly..

tan² x - sin² x = tan² x sin² x
1(sin² x)/(cos² x) - sin² x
2(sin² x)/(cos² x) - (sin² x cos² x)/(cos² x)
3(sin² x - sin² x cos² x)/(cos² x)
4(sin² x (1 - cos² x))/(cos² x)
5(sin² x (sin² x))/(cos² x)
6(sin² x sin² x)/ (cos² x)


ok after line 6 i am a little stuck, should i break up the fraction into two?

I suspect that you chose not to look at the right hand side of the equation because if you did you would not have even seen one fraction none-the-less two. How do you see two fractions. A fraction has a denominator! On factor on the rhs has sin^2 x, just like you have on the LHS. The only other factor the RHS is Tan^2 x. The only factor left on the LHS is (sin² x)/ (cos² x).
At this point we have only two things going on: Either (sin² x)/ (cos² x)= tan² x or you have a mistake somewhere. Which is it??

 

[abel muroi]

I suspect that you chose not to look at the right hand side of the equation because if you did you would not have even seen one fraction none-the-less two. How do you see two fractions. A fraction has a denominator! On factor on the rhs has sin^2 x, just like you have on the LHS. The only other factor the RHS is Tan^2 x. The only factor left on the LHS is (sin² x)/ (cos² x).
At this point we have only two things going on: Either (sin² x)/ (cos² x)= tan² x or you have a mistake somewhere. Which is it??

so the right hand side is also (sin² x sin² x)/(cos² x)?

 

[Steven G]

so the right hand side is also (sin² x sin² x)/(cos² x)?

Basically yes. You need to realize that ALGEBRA (here we go again) allows us to write (ab)/c as (a/c)*b

(sin² x sin² x)/(cos² x)=sin² x (sin² x/cos² x) = (sin² x tan² x). Done

Learn algebra. The trig is hard enough without having algebra trouble-like factoring and the concept above