Question 604040
{{{1/(cot(x)+1) + 1/(tan(x)+1) = 1}}}
When trying to do these identities, one of the things to aim for is to match the number of terms on each side of the equation. Your equation has just two terms on the left and it has one term on the right. So somehow we need to condense the left side from two terms into one.<br>
Changing the number of terms is often done by one of the following:<ul><li>Using a Trig property that has a different number of terms on the two sides. For example, {{{tan^2(x) + 1 = sec^2(x)}}}</li><li>Adding or subtracting fractions. For example: {{{(a+b)/c = a/c + b/c}}} or {{{(a-b)/c = a/c - b/c}}}</li></ul>
Since your left side already has fractions, let's go ahead and add them. Of course we need a common denominator, first:
{{{1/(cot(x)+1) + 1/(tan(x)+1) = 1}}}
{{{(1/(cot(x)+1))((tan(x)+1)/(tan(x)+1)) + (1/(tan(x)+1))((cot(x)+1)/(cot(x)+1)) = 1}}}
which gives us:
{{{(tan(x)+1)/((cot(x)+1)*(tan(x)+1)) + (cot(x)+1)/((tan(x)+1)*(cot(x)+1)) = 1}}}
Using FOIL to multiply out the denominator we get:
{{{(tan(x)+1)/(1 + cot(x) + tan(x) + 1) + (cot(x)+1)/(1 + cot(x) + tan(x) + 1) = 1}}}
Adding the fractions we get:
{{{(tan(x)+1 + cot(x)+1)/(1 + cot(x) + tan(x) + 1) = 1}}}
The numerator and denominator are equal are equal and dividing something by itself (except for zero) is always equal to a 1:
1 = 1
And we're done!