SOLUTION: how do you solve the equation that reads as follows: (cos theta over 1+sin theta) + (1+sin theta over cos theta) = 2sec theta ?

Question 216754: how do you solve the equation that reads as follows: (cos theta over 1+sin theta) + (1+sin theta over cos theta) = 2sec theta ?
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
(Algebra.com's software makes it difficult to display theta correctly. So I will be using "x" instead of theta. Just replace all the x's in the solution below with theta's.)

Often it makes things easier in Trig. if you rewrite everything in terms of sin and/or cos. So we'll start by replacing sec(x) with 1/cos(x):

Also, the fewer fractions an equation has, the easier it is to work with. Since 2 of the 3 fractions have the same denominator, we can eliminate those two by multiplying both sides of the equation by cos(x):

Using the distributive property on the left side we get:

Now when we multiply the cos(x)'s cancel in the second and third fractions:

We still have one fraction. The easiest way to handle it is to see what happens if we use the fact that . Substituting we get:

Now we can eliminate the fraction by factoring and reducing!

The (1 + sin(x))'s cancel leaving:

Adding like terms we get:

What does this equation mean? What happened to the x's (or theta's)? The fact that the x's are gone means that the value of x is irrelevant. The original equation is either always true or never true, regardless of what value x (or theta) has. The fact that the equation we ended up with, 2=2, was true means that all possible values of x make the original equation true.
An equation like this that is true for all possible values of x is called an "identity". So we have shown that

is an identity.

(Note the use of the phrase "possible values of x". This is a reference to the domain. The original equation is undefined for all odd multiples of 90 degress because they make one or more of the denominators zero. Therefore the domain is all numbers except the odd multiples of 90 and the equation is true for all numbers except these odd multiples of 90.)

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