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

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

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.

\Large \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