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orkdork · Oct 25, 2009, 05:44 PM

I tried this proof all day and I didn't get the right answer... :mad:

tanx - cotx+ (2cos^2 x) / (cosx sinx) = secx cscx

Perito · Oct 25, 2009, 07:35 PM

Originally Posted by orkdork

I tried this proof all day and I didn't get the right answer.

tanx - cotx+ (2cos^2 x) / (cosx sinx) = secx cscx

Is this

$tan(x) - cot(x) + \frac {2cos^2(x)}{cos(x)sin(x)} = sec(x)csc(x)$

or

$\frac {tan(x) - cot(x) + 2cos^2(x)}{cos(x)sin(x)} = sec(x)csc(x)$

I think you mean the first.

You should memorize the following (they're used in a lot of proofs)

$tan(x) = \frac {sin(x)}{cos(x)}$

$cot(x) = \frac {1}{tan(x)} = \frac {cos(x)}{sin(x)}$

$sec(x) = \frac {1}{cos(x)}$

$csc(x) = \frac {1}{sin(x)}$

$cos^2(x) + sin^2(x) = 1$

So, let's try substituting.

$\frac {sin(x)}{cos(x)} - \frac {cos(x)}{sin(x)}+ \frac {2cos^2(x))}{cos(x) sin(x)}= \frac {1}{cos(x)} \frac {1}{sin(x)}$

Bry putting everything over a common denominator [sin(x)cos(x)]

$\frac {sin^2(x) - cos^2(x) + 2cos^2(x)}{sin(x)cos(x)} = \frac {1}{sin(x)cos(x)}$

$\frac {[sin^2(x) +cos^2(x)] + [cos^2(x) -cos^2(x) }{sin(x)cos(x)} = \frac {1}{sin(x)cos(x)}$

$\frac {[1] + [0]}{sin(x)cos(x)} = \frac {1}{sin(x)cos(x)}$

and

$\frac {1}{sin(x)cos(x)} = \frac {1}{sin(x)cos(x)}$

QED