Question 837869
Note 1: To save on typing I'm going to use x instead of theta.
Note 2: Algebra.com's formula drawing software does not handle csc or sec or cot well. Sometimes it looks like there is a multiplication symbol between the function and its argument. There is no multiplication involved there so please ignore the symbols.<br>
{{{((csc(x)-1)/csc(x))*((csc(x)+1)/csc(x))=1/sec^2(x)}}}
One way is to approach it straightforward. We'll go ahead and multiply the two fractions on the left side. (In the numerators we can use the {{{(a-b)(a+b) = a^2-b^2}}} pattern to multiply quickly.)
{{{(csc^2(x)-1)/csc^2(x)=1/sec^2(x)}}}<br>
Our next step could be several things. We will separate the fraction into two (because it will simplify easily that way):
{{{csc^2(x)/csc^2(x)-1/csc^2(x)=1/sec^2(x)}}}
which simplifies to:
{{{1-1/csc^2(x)=1/sec^2(x)}}}<br>
Since csc and sin are reciprocals of each other, the second fraction can be replaced:
{{{1-sin^2(x)=1/sec^2(x)}}}<br>
We should recgonize the left side as one of the Pythagorean identities. So we can replace it:
{{{cos^2(x)=1/sec^2(x)}}}
And since cos and sec are reciprocals of each other, too:
{{{1/sec^2(x)=1/sec^2(x)}}}<br>
P.S. If you have trouble with these types of problems, then something to try when you can't think of anything else is to rewrite any sec's, csc's, tan's and cot's as sin's and/or cos's. Here's another way to solve your problem which uses this idea:
{{{((csc(x)-1)/csc(x))*((csc(x)+1)/csc(x))=1/sec^2(x)}}}
{{{((1/sin(x)-1)/(1/sin(x)))((1/sin(x)+1)/(1/sin(x)))=1/sec^2(x)}}}
This may look pretty ugly. But it will simplify easily. You may remember from algebra that fractions within a fraction can be simplified by multiplying the numerator and denominator of the "big" fraction by the lowest common denominator (LCD) of the "little" fractions. Fortunately all the "little" denominators are the same so the LCD is simple and asy to find.
{{{(((1/sin(x)-1)/(1/sin(x)))(sin(x)/sin(x)))(((1/sin(x)+1)/(1/sin(x)))(sin(x)/sin(x))))=1/sec^2(x)}}}
which simplifies to:
{{{((1-sin(x))/1)((1+sin(x))/1)=1/sec^2(x)}}}
which simplifies further to:
{{{1-sin^2(x) = 1/sec^2(x)}}}
Using the Pythagorean identity the left side becomes:
{{{cos^2(x)=1/sec^2(x)}}}
And since cos and sec are reciprocals of each other:
{{{1/sec^2(x)=1/sec^2(x)}}}