Question 223721
(Algebra'com's formula software doesn't "do" theta's for some reason. So I will use t instead.)
Since this problem doesn't work out very well as you've written it, I going to assume that there are supposed to be parentheses:
{{{(csc(t) + cot(t))(csc(t) - cot(t)) = 1}}}
If this is wrong, stop reading.<br>
When you don't see a better way to solve a problem in Trig., it is often helpful to rewrite the expressions in terms of sin and cos. So we will be replacing csc(t) with 1/sin(t) and cot(t) with cos(t)/sin(t):
{{{(1/sin(t) + cos(t)/sin(t))*(1/sin(t) - cos(t)/sin(t)) = 1}}}
To multiply this it helps if you recognize that it fits the pattern: {{{(a+b)(a-b) = a^2 - b^2}}} with "a" being 1/sin(t) and "b" being cos(t)/sin(t). Using this pattern (or multiplying out the long way with FOIL and combining like terms) we get:
{{{(1/sin(t))^2 - (cos(t)/sin(t))^2 = 1}}}
which simplifies to:
{{{1/(sin(t))^2 - (cos(t))^2/(sin(t))^2 = 1}}}
The fractions have the same denominator so we can subtract them:
{{{(1 - (cos(t))^2)/(sin(t))^2 = 1}}}
Since {{{1 - (cos(t))^2 = (sin(t))^2}}} the numerator becomes:
{{{(sin(t))^2/(sin(t))^2 = 1}}}
And the fraction cancels leaving
{{{1 = 1}}}
And with this final equation, since 1 always equal 1, shows that your original equation is also always true. We call this an identity.