# Verifying trig identities

• Oct 28th 2009, 03:34 PM
l flipboi l
Verifying trig identities
Hello,

Problem: (tan-cot) / (tan^2) - (cot^2) = sin cos
if I'm right the problem should look like this...

((sin/cos)-(cos/sin)) / ((sin^2/cos^2)-(cos^2/sin^2))

what are the algebraic steps to verify this problem?

Thanks!
• Oct 28th 2009, 03:39 PM
e^(i*pi)
Quote:

Originally Posted by l flipboi l
Hello,

Problem: (tan-cot) / (tan^2) - (cot^2) = sin cos
if I'm right the problem should look like this...

$((sinA/cosA)-(cosA/sinA)) / ((sin^2A/cos^2A)-(cos^2A/sin^2A))$

what are the algebraic steps to verify this problem?

Thanks!

Yep, so far so good. I would then cross multiply to give only one denominator:

$\frac{sinA}{cosA} - \frac{cosA}{sinA} = \frac{sin^2A-cos^2A}{sinAcosA}$

$\frac{sin^2A}{cos^2A} - \frac{cos^2A}{sin^2A} = \frac{sin^4A-cos^4A}{sin^2Acos^2A}$

Therefore we get:

$\frac{sin^2A-cos^2A}{sinAcosA} \times \frac{sin^2Acos^2A}{sin^4A-cos^4A}$.

To remove the quartic recall the difference of two squares and that $sin^2A+cos^2A = 1$
• Oct 29th 2009, 03:09 PM
l flipboi l
Quote:

Originally Posted by e^(i*pi)
Yep, so far so good. I would then cross multiply to give only one denominator:

$\frac{sinA}{cosA} - \frac{cosA}{sinA} = \frac{sin^2A-cos^2A}{sinAcosA}$

$\frac{sin^2A}{cos^2A} - \frac{cos^2A}{sin^2A} = \frac{sin^4A-cos^4A}{sin^2Acos^2A}$

Therefore we get:

$\frac{sin^2A-cos^2A}{sinAcosA} \times \frac{sin^2Acos^2A}{sin^4A-cos^4A}$.

To remove the quartic recall the difference of two squares and that $sin^2A+cos^2A = 1$

Thanks for the help! I was having difficulties with dividing fractions