# Trig Identities - where is my error?

• Oct 2nd 2009, 03:09 AM
Xplorer
Trig Identities - where is my error?
Prove:
$csc(A)^2 + sec(A)^2 = csc(A)^2sec(A)^2$

left side is:
$(1/cos(A)^2) + (1/sin(A)^2)$

I multiply the first term by $(sin(A)^2)/(sin(A)^2)$
and the second term by $(cos(A)^2)/(cos(A)^2)$

$(sin(A)^2)/(sin(A)^2)(1/cos(A)^2) + (cos(A)^2)/(cos(A)^2)(1/sin(A)^2)$

Now the two terms have a common denominator and can be combined and we have:
$(sin(A)^2 + cos(A)^2)/(2cos(A)^2sin(A)^2)$

Then numerator = 1:
$1/(2cos(A)^2sin(A)^2)$

Then we have
$(1/2)csc(A)^2sec(A)^2$

Where does the $1/2$ come from? What am I doing wrong? How can I make the left side look like $csc(A)^2sec(A)^2$ ???

Thanx

-Tony
• Oct 2nd 2009, 03:25 AM
Hello Tony

Welcome to Math Help Forum!
Quote:

Originally Posted by Xplorer
...

$(sin(A)^2)/(sin(A)^2)(1/cos(A)^2) + (cos(A)^2)/(cos(A)^2)(1/sin(A)^2)$

Now the two terms have a common denominator and can be combined and we have:
$(sin(A)^2 + cos(A)^2)/(2cos(A)^2sin(A)^2)$

Ouch! Since when have you added fractions like that?

$\frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}$,
not $\frac{a+c}{2b}$.

• Oct 2nd 2009, 04:26 PM
Xplorer
Oh my gosh!!! Now I see it. It was late at night when I was doing that. Wow I feel stupid now (Itwasntme)

Thank you very much. lol!

-Tony