Hi, there--

YOUR PROBLEM:
Prove that (sin x + cos x)(tan x + cot x) = sec x + csc x

A SOLUTION:
In order to prove a trigonometric identity, we work on one side of the equation, rewriting it 
as a series of equivalent expressions until both sides of the equation are identical.

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

Let's work on the left hand side. Apply the distributive property.
(sin x)(tan x) + (sin x)(cot x) + (cos x)(tan x) + (cos x)(cot x) = sec x + csc x

Write tan x and cot x in terms of sin x and cos x
(sin x)(sin x)/(cos x) + (sin x)(cos x)/(sin x) + (cos x)(sin x) /(cos x) + (cos x)(cos x) /(sin x) 
= sec x + csc x

Simplify, canceling terms where possible.
(sin x)^2/(cos x) + (cos x) + (sin x) + (cos x)^2/(sin x) = sec x + csc x

Rewrite each term with a denominator of (sin x)( cos x).
[(sin x)^2(sin x)]/[(sin x)(cos x)] + [(cos x)^2(sin x)]/[(sin x)(cos x)] + [(sin x)^2(cos x)]/[(sin x)
(cos x)] + [(cos x)^2(cos x)]/[(sin x)(cos x)] = sec x + csc x

Factor sin x from the 1st and 2nd terms and cos x from the 3rd and 4th terms.
(sin x)[(sin x)^2 + (cos x)^2)]/[(sin x)(cos x)] + (sin x)[(sin x)^2 + (cos x)^2)]/[(sin x)(cos x)] 
= sec x + csc x

Use the trig identify, (sin x)^2 + (cos x)^2 = 1, to simplify. Cancel like terms in numerator 
and denominator.
([sin x + cos x]/[(sin x)(cos x)] = sec x + csc x
(sin x)/[(sin x)(cos x)] + (cos x)/[(sin x)(cos x)] = sec x + csc x
1/(cos x) + 1/(sin x) = sec x + csc x

Use the definition of sec x and csc x to rewrite the equation.
sec x + csc x = sec x + sec x

That's it.

Hope that helps,
Mrs. Figgy