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!