Stuck verifying: (tan^2 x)+(cot^2 x) = (tan x + cot x )^2 - 2

[JamesMp]

Like the title said, I have been stuck trying to verify this: tan²x+cot²x = (tanx+cotx)²-2 for quite some time. Been searching around for something to point me in the right direction but so far have come up blank. If someone can shed some light on this for me I would be oh so grateful. Thanks!

 

[stapel]

I have been stuck trying to verify this:

tan²x+cot²x = (tanx+cotx)²-2 for quite some time.

Please reply showing at least one of the attempts you have made during this lengthy time period. (Maybe you're almost to the solution!) Thank you!

 

[JamesMp]

Please reply showing at least one of the attempts you have made during this lengthy time period. (Maybe you're almost to the solution!) Thank you!

I don't think I'm very close, but I can see on the right side that (tanx+cotx)^2 would be equal to tan^2x+2tanxcotx+cot^2 and I'm assuming the 2 coefficient is where the 2 in the - 2 is coming from if that makes sense. Although now that I think about that doesn't really make sense at all.
I've also tried setting it up like (sin^2x)/(cos^2x) + (cos^2x)/(sin^2x) and then getting a common denominator but I'm not sure where to go from there.
I also tried setting it up (1-sec^2x)+(1-csc^2) but I again couldn't think of anywhere to go from there and have a feeling that's not part of the solution.

Thanks for any help...

 

[Deleted member 4993]

I don't think I'm very close, but I can see on the right side that (tanx+cotx)^2 would be equal to tan^2x+2tanxcotx+cot^2 and I'm assuming the 2 coefficient is where the 2 in the - 2 is coming from if that makes sense. Although now that I think about that doesn't really make sense at all.
I've also tried setting it up like (sin^2x)/(cos^2x) + (cos^2x)/(sin^2x) and then getting a common denominator but I'm not sure where to go from there.
I also tried setting it up (1-sec^2x)+(1-csc^2) but I again couldn't think of anywhere to go from there and have a feeling that's not part of the solution.

Thanks for any help...

You got:

[tan(x)+cot(x)]^2 = tan^2(x) + 2 * tan(x) * cot(x) + cot^2(x)

You are very close...

Consider the fact that

$\displaystyle \tan (x) = \dfrac{1}{\cot(x)}$

In light of that:

tan(x) * cot(x) = ?

 

[JamesMp]

Right, so they cancel each other out leaving tan^2x+2+cot^2x
But I'm just not seeing where to go from there.

Thanks

 

[HallsofIvy]

What do you mean "go from there"? You were asked to prove that $\displaystyle (tan(x)+ cot(x))^2- 2= tan^2(x)+ cot^2(x)$.

You have been shown how to change the left side to $\displaystyle tan^2(x)+ cot^2(x)$. That's all you need to do!

 

[JamesMp]

What do you mean "go from there"? You were asked to prove that $\displaystyle (tan(x)+ cot(x))^2- 2= tan^2(x)+ cot^2(x)$.

You have been shown how to change the left side to $\displaystyle tan^2(x)+ cot^2(x)$. That's all you need to do!

Thanks, I do see that, I guess what had me confused is that I was trying to turn tan^2(x)+cot^2(x) into (tanx+cotx)^2-2 rather than the other way around. Thanks for the help.

 

[HallsofIvy]

Thanks, I do see that, I guess what had me confused is that I was trying to turn tan^2(x)+cot^2(x) into (tanx+cotx)^2-2 rather than the other way around. Thanks for the help.

Well, if you really want to the tan^2(x)+ cot^2(x)= tan^2(x)+ 2+ cot^2(x)- 2= tan^2(x)+ 2(tan(x))(cot(x))+ cot^2(x)- 2= (tan(x)+ cot(x))^2- 2.