Trigonometry proof... help please

[math.]

The problem: Prove that tan²a+1 = sec²a

 

[pka]

The problem: Prove that tan²a+1 = sec²a

Can you show that \(\displaystyle \tan^2(a)+1=\dfrac{\sin^2(a)+\cos^2(a)}{\cos^2(a)}~?\)

 

[math.]

Can you show that \(\displaystyle \dfrac{\sin^2(a)+\cos^2(a)}{\cos^2(a)}~?\)

nope, If you change the sec^2 a to \(\displaystyle \dfrac{\sin^2(a)+\cos^2(a)}{\cos^2(a)}\) where does the cos² in the numerator come from?

 

[Deleted member 4993]

nope

try

\(\displaystyle tan(a) \ = \ \dfrac{sin(a)}{cos(a)}\)

 

[JeffM]

nope

Wow. This effort leaves me breathless.

 

[math.]

try

\(\displaystyle tan(a) \ = \ \dfrac{sin(a)}{cos(a)}\)

In the problem it gives that as a hint, but then I wind up with sec²a = \(\displaystyle \dfrac{sin^2(a)}{cos^2(a)}\) and I'm not sure about where to go from there.

 

[Deleted member 4993]

In the problem it gives that as a hint, but then I wind up with sec²a = \(\displaystyle \dfrac{sin^2(a)}{cos^2(a)}\) and I'm not sure about where to go from there.

How did you get that?

tan²a = \(\displaystyle \dfrac{sin^2(a)}{cos^2(a)}\)

Now add 1 to this fraction.

 

[math.]

How did you get that?

tan²a = \(\displaystyle \dfrac{sin^2(a)}{cos^2(a)}\)

Now add 1 to this fraction.

adding 1 : \(\displaystyle \dfrac{sin^2(a)}{cos^2(a)}\) + 1. Where am I supposed to go from there?

 

[Deleted member 4993]

adding 1 : \(\displaystyle \dfrac{sin^2(a)}{cos^2(a)}\) + 1. Where am I supposed to go from there?

\(\displaystyle \dfrac{sin^2(a)}{cos^2(a)} + 1 \ = \ \dfrac{sin^2(a)}{cos^2(a)} + \dfrac{cos^2(a)}{cos^2(a)} \)

 

[math.]

\(\displaystyle \dfrac{sin^2(a)}{cos^2(a)} + 1 \ = \ \dfrac{sin^2(a)}{cos^2(a)} + \dfrac{cos^2(a)}{cos^2(a)} \)

If you add one then wouldn't you end up with \(\displaystyle \dfrac{sin^2(a) + cos^2(a)}{2cos^2(a)} \)... which has two of cos²(a) in the denominator whereas the one that pka posted only has one.

 

[Deleted member 4993]

If you add one then wouldn't you end up with \(\displaystyle \dfrac{sin^2(a) + cos^2(a)}{2cos^2(a)} \)... which has two of cos²(a) in the denominator whereas the one that pka posted only has one.

How much is:

\(\displaystyle \dfrac{1}{2} + \dfrac{1}{2}\)

certainly not \(\displaystyle \dfrac{2}{4} \) !!!

 

[math.]

Of course not.... so from \(\displaystyle \dfrac{sin^2(a) + cos^2(a)}{cos^2(a)} \) = sec²a where do I go?

 

[JeffM]

Of course not.... so from \(\displaystyle \dfrac{sin^2(a) + cos^2(a)}{cos^2(a)} \) = sec²a where do I go?

\(\displaystyle tan^2(\alpha) + 1 = \left(\dfrac{sin(\alpha)}{cos(\alpha)}\right)^2 + 1 = \dfrac{sin^2(\alpha)}{cos^2(\alpha)} + 1 = \dfrac{sin^2(\alpha) + cos^2(\alpha)}{cos^2(\alpha)} = \dfrac{1}{cos^2(\alpha)}\)

 

[math.]

\(\displaystyle tan^2(\alpha) + 1 = \left(\dfrac{sin(\alpha)}{cos(\alpha)}\right)^2 + 1 = \dfrac{sin^2(\alpha)}{cos^2(\alpha)} + 1 = \dfrac{sin^2(\alpha) + cos^2(\alpha)}{cos^2(\alpha)} = \dfrac{1}{cos^2(\alpha)}\)

and then \(\displaystyle \dfrac{1}{cos^2(\alpha)}\) = sec²(a). Cool, thanks all.