Verifying trig identities

I'm stuck on a problem at the moment was just going to see if anyone could tell me how they would work it.

1+sinx divided by cosx + cosx divided by 1 + sinx = 2secx

Generally, when you are adding fractions, you should try and combine them.
the common factor in this case is cos(x) * (1 + sin(x) ), so multiply the left fraction by (1 + sin(x) )/ (1 + sin(x) ) and multiply the right fraction by cos(x) / cos(x)

((1+sin(x) )(1+sin(x) ) + cos(x)^2)/ ( cos(x) * (1 + sin(x) ) )

This looks messy, but its not as bad as it seems. First distribute the 1 +sin(x)

( 1 + 2sin(x) + sin(x) ^2 + cos(x) ^2 ) / (cos(x) * (1 + sin(x) ) )

We know that sin(x)^2 + cos(x)^2 = 1, so substitute

2 + 2sin(x) / (cos(x) * (1 + sin(x) ) )

we factor out the 2, get (1 + sin(x) ) in the denominator, so we can cancel the (1 +sin(x) ) term on the numerator and in the denominator.

2/cos(x)

Which we know equals 2sec(x).

Voila! Hope that helps. Combining fractions