How do you prove $cot(theta)sec(theta)=csc(theta)$ ?

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Answer 1 · 2016-05-30T18:27:57

Applying the definition of $cot, sec, csc$ and $tan$ .

Recall the definition of these functions:

$cot(theta)=1/tan(theta)$
$sec(theta)=1/cos(theta)$
$csc(theta)=1/sin(theta)$

Then we want to prove

$cot(theta)sec(theta)=csc(theta)$

that is equivalent to

$1/tan(theta)1/cos(theta)=1/sin(theta)$

We recall that $tan(theta)=sin(theta)/cos(theta)$ , consequently
$1/tan(theta)=cos(theta)/sin(theta)$ .
I substitute in the previous equation

$1/tan(theta)1/cos(theta)=1/sin(theta)$

$cos(theta)/sin(theta)1/cos(theta)=1/sin(theta)$

$1/sin(theta)=1/sin(theta)$ .

How do you prove cot(theta)sec(theta)=csc(theta)cot(theta)sec(theta)=csc(theta)?