Verify the Identity?

[Earl95]

Hi guys,

Could I get some help with a little Trig work? I have to verify the identity:
tan^(2)xcos^(2)x+cot^(2)xsin^(2)x=1.

So far I have
(1-sec^(2)x)cos^2x+(1-csc^(2)x)sin^2x
cos^(2)x-cos^(2)xsec^2+sin^(2)x-sin^(2)xcsc^(2)x
cos^(2)x-cos^(2)x(1/(cos^(2)x))+sin^(2)x-sin^(2)x(1/(sin^(2)x)

and I'm stumped. Am I even doing it right? Is the work so far so good, or is it wrong?

Any help is greatly appreciated!
--Earl

 

[BigGlenntheHeavy]

\(\displaystyle tan^2(x)cos^2(x)+cot^2(x)sin^2(x) \ = \ 1\)

\(\displaystyle =\bigg(\frac{sin^2(x)}{cos^2(x)}\bigg)\bigg(\frac{cos^2(x)}{1}\bigg)+\bigg(\frac{cos^2(x)}{sin^2(x)}\bigg)\bigg(\frac{sin^2(x)}{1}\bigg) \ = \ 1\)

\(\displaystyle = \ sin^2(x)+cos^2(x) \ = \ 1, \ 1 \ = \ 1, \ QED\)

 

[Earl95]

\(\displaystyle tan^2(x)cos^2(x)+cot^2(x)sin^2(x) \ = \ 1\)

\(\displaystyle =\bigg(\frac{sin^2(x)}{cos^2(x)}\bigg)\bigg(\frac{cos^2(x)}{1}\bigg)+\bigg(\frac{cos^2(x)}{sin^2(x)}\bigg)\bigg(\frac{sin^2(x)}{1}\bigg) \ = \ 1\)

\(\displaystyle = \ sin^2(x)+cos^2(x) \ = \ 1, \ 1 \ = \ 1, \ QED\)

Thanks! Is that the only way to solve it, though, because the question asks if the way that I posted was correct or not. I don't know how to tell.

 

[Deleted member 4993]

\(\displaystyle tan^2(x)cos^2(x)+cot^2(x)sin^2(x) \ = \ 1\)

\(\displaystyle =\bigg(\frac{sin^2(x)}{cos^2(x)}\bigg)\bigg(\frac{cos^2(x)}{1}\bigg)+\bigg(\frac{cos^2(x)}{sin^2(x)}\bigg)\bigg(\frac{sin^2(x)}{1}\bigg) \ = \ 1\)

\(\displaystyle = \ sin^2(x)+cos^2(x) \ = \ 1, \ 1 \ = \ 1, \ QED\)

Thanks! Is that the only way to solve it, though, because the question asks if the way that I posted was correct or not. I don't know how to tell.

There are several different ways - but more complicated - to prove this.

Once you got the LHS = RHS ? you are most probably correct.

 

[BigGlenntheHeavy]

\(\displaystyle An \ Identity \ equals \ itself, \ for \ example, \ (x-2)(x+3) \ = \ x^2+x-6.\)

\(\displaystyle In \ other \ words, \ no \ matter \what \ number \ you \ let \ x \ equal, \ both \ sides \ of\)

\(\displaystyle the \ equation \ will \ be \ equal, \ try \ it, \ whereas \ a \ conditional \ equation \ like \ x-3 \ = \ 0\)

\(\displaystyle has \ only \ one \ solution, \ namely \ x \ = \ 3.\)

 

[Earl95]

Hi guys,

Could I get some help with a little Trig work? I have to verify the identity:
tan^(2)xcos^(2)x+cot^(2)xsin^(2)x=1.

So far I have
(1-sec^(2)x)cos^2x+(1-csc^(2)x)sin^2x
cos^(2)x-cos^(2)xsec^2+sin^(2)x-sin^(2)xcsc^(2)x
cos^(2)x-cos^(2)x(1/(cos^(2)x))+sin^(2)x-sin^(2)x(1/(sin^(2)x)
cos^(2)x+sin^(2)=1
1=1

and I'm stumped. Am I even doing it right? Is the work so far so good, or is it wrong?

Any help is greatly appreciated!
--Earl

So, what I'm really asking is: is this way correct? I understand that there are other ways, but I need to know if this way is mathematically true.

 

[wjm11]

is this way correct? I understand that there are other ways, but I need to know if this way is mathematically true.

Yes, your method is fine. In working with trig identities, there are usually multiple ways to go.

 

[Earl95]

Thank you all for helping me out. I'm all good now.