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.
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!