# How to verify or prove this identiy?Here is the identity I need to prove: (sin x + cos x)(tan x + cot x) = sec x + csc x Thanks to any answer!

jellybean1977 | Certified Educator

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.

justaguide | Certified Educator

We have to prove that (sin x + cos x)(tan x + cot x) = sec x + csc x

(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