Answer:
Step-by-step explanation:
It is actually easier to ignore the left side for now, and work with the right side. The secant function can be represented in terms of cosine by putting a one over cosine. This relation below should be known:
[tex]sec(x)=\frac{1}{cos(x)}[/tex]
So, if we replace the secants on the right with the above relation, you can make the right side in terms completely of cosine.
[tex]\frac{1-cos(x)}{1+cos(x)} =\frac{\frac{1}{cos(x)}-1 }{\frac{1}{cos(x)} +1}[/tex]
I hope that's clear to see. Brainly equations are pretty small and blurry.
The next step is to get a common denominator for the right side. The term '1' is equivalent to cos(x)/cos(x). So we have the following:
[tex]\frac{1-cos(x)}{1+cos(x)} =\frac{\frac{1}{cos(x)}-\frac{cos(x)}{cos(x)} }{\frac{1}{cos(x)} +\frac{cos(x)}{cos(x)} }[/tex]
Looks rather complex, but it simplifies everything by a lot. Combine it into one single fraction:
[tex]\frac{1-cos(x)}{1+cos(x)} =\frac{\frac{1-cos(x)}{cos(x)} }{\frac{1+cos(x)}{cos(x)} }[/tex]
When you divide by a fraction, it is equivalent to multiplying by the reciprocal. This is an algebra 1 concept, so it should be familiar:
[tex]\frac{1-cos(x)}{1+cos(x)} =\frac{1-cos(x)}{cos(x)}*\frac{cos(x)}{1+cos(x)}[/tex]
As you can see, the cos(x)'s will cancel, leaving:
[tex]\frac{1-cos(x)}{1+cos(x)} =\frac{1-cos(x)}{1+cos(x)}[/tex]
QED
