We have to prove that (sin x + cos x)(tan x + cot x) = sec x + csc x
Let's start with the left hand side
(sin x + cos x)(tan x + cot x)
=> (sin x + cos x)*[(sin x/cos x) + (cos x/sin x)]
=> (sin x + cos x)*[(sin x)^2 + (cos x)^2]/(cos x*sin x)
=> (sin x + cos x)/(cos x * sin x)
=> sin x/(cos x * sin x) + cos x/(cos x * sin x)
=> 1/cos x + 1/sin x
=> sec x + csc x
which is the right hand side
This proves that (sin x + cos x)(tan x + cot x) = sec x + csc x
First you need to remember that tan x is the same as sin x/cos x, cot x = cos x/sin x, sec x = 1/cos x, and csc x = 1/sin x.
Then if you look at the left side of the equation and multiple out the terms using FOIL, you'll get:
sin^2 x/cos x + cos x + sin x + cos^2 x/sin x = sec x + csc x
Since you have fractions on the left side, and you are adding, you need common denominators, so I changed cos x to cos^2 x/cos x and sin x to sin^2 x/sin x. Then you can add the two fractions with cos x in the denominator and the two fractions with the sin x in the denominator, this gives you:
(sin^2 x +cos^2 x)/cos x + (sin^2 x + cos^2 x)/sin x = sec x + csc x
The last thing you need to remember is the identity: sin^2 x + cos^2 x = 1, so you can replace both of the numerators with one and this gives you:
1/cos x + 1/sin x = sec x + csc x
Now, by the definitions of secant and cosecant that I listed at the top, that proves the identity.
Hopefully that helps.