tan(x)/sec(x)+1=csc(x)-cot(x)

tan(x)/sec(x) + 1 does not equal csc(x)-cot(x)

I think what you mean is probably this:

tan(x)/(sec(x)+1) = csc(x)-cot(x)

As usual for these problems - convert the trig functions to sine and cosone equivalents, and simplify.

I'm not able to simplify. What I had is correct. I need to figure out how to get one side to equal the other. And I don't know how to do this.

First convert the tan(x), sec(x), csc(x) and cot(x) functions to sin(x) and cos(x) equivalents:

tan(x) = sin(x)/cos(x)
cot(x) = cos(x)/sin(x)
sec(x) = 1/cos(x)
csc(x) = 1/sin(x)

So for the eleft hand side you have:


\frac{ tan(x)}{sec(x)+1} = \frac {\sin(x)/\cos(x)}{1/\cos(x) +1}


Then multiply through by cos(x):


\frac {\sin(x)/\cos(x)}{1/\cos(x) +1} \times \frac {\cos(x)}{\cos(x)} = \frac {\sin(x)}{1 + \cos(x)}


Now do the same technique to the right hand side. The final step will be to cross-multiply the numerator of the left hand side by the denominator of the right, and the numerator of the right by the denominator of the left, then show they are equivalent.