# I could use some help proving that `(tanx+cotx)/(secx+cscx)=1/(cosx+sinx)`I have worked this problem from both sides several times and keep coming up with unmatched answers.

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Probably the best way to start here is to convert everything in the first part of the equation into sines and cosines. Remember the following identities:

`tanx = sinx/cosx`

`cotx = 1/tanx = cosx/sinx`

`secx = 1/cosx`

`cscx = 1/sinx`

Using these, we can make the following substitution:

`(tanx+cotx)/(secx+cscx) = (sinx/cosx + cosx/sinx)/(1/cosx+1/sinx)`

To simplify this, let's multiply the top and bottom by `sinxcosx`:

`((sinx/cosx+cosx/sinx)sinxcosx)/((1/cosx+1/sinx)sinxcosx) = (sin^2x + cos^2x)/(sinx + cosx)`

Notice that the distributive property will give us the above result. Now, to complete the problem, you simply need to recall the following trigonometric identity:

`sin^2x + cos^2x = 1`

Remember, this identity comes about because `cosx` and `sinx` are the `x` and `y` coordinates, respectively, of any point on the unit circle or simply by their own respective definitions!

Therefore, we get the following substitution:

`(sin^2x + cos^2x)/(sinx + cosx) = 1/(sinx + cosx)`

There is your desired result! I hope that helps!