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Calculus Student · Dec 14, 2007, 01:50 AM

Usually I am pretty good at these, but this one is killing me. Please help!

Never mind I solved it. Here's the answer if anyone is curious.

$\frac{sin^2\theta-tan\theta}{cos^2\theta-cot\theta}=tan^2\theta$

$\frac{sin^2\theta-\frac{sin\theta}{cos\theta}}{cos^2\theta-\frac{cos\theta}{sin\theta}}$

$\frac{\frac{sin^2\theta cos\theta-sin\theta}{cos\theta}}{\frac{cos^2\theta sin\theta-cos\theta}{sin\theta}}$

$\frac{\frac{sin\theta(sin\theta cos\theta-1)}{cos\theta}}{\frac{cos\theta(sin\theta cos\theta-1)}{sin\theta}}$

$(\frac{sin\theta(sin\theta cos\theta-1)}{cos\theta}) X (\frac{sin\theta}{cos\theta(sin\theta cos\theta-1)})$

$\frac{(sin\theta)(sin\theta)(sin\theta cos\theta-1)}{(cos\theta)(cos\theta)(sin\theta cos\theta-1)}$

$\frac{(sin\theta)(sin\theta)\cancel{(sin\the ta cos\theta-1)}}{(cos\theta)(cos\theta)\cancel{(sin\theta cos\theta-1)}}$

$\frac{sin^2\theta}{cos^2\theta}$

$\frac{sin^2\theta}{cos^2\theta}=tan^2\theta$

ISneezeFunny · Dec 14, 2007, 02:54 AM

it's 5am here... so I may be a bit off. I'll get you started:

tan = sin/cos
cot = cos/sin

numerator: sin^2 - sin/cos
denominator: cos^2 - cos/sin

take out sin from num, cos from den

you end up with

sin(sin - 1/cos)
==========
cos(cos - 1/sin)

so you can get 1 TAN out of that. Hope that gets you started.