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

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

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Answer 1 · 2016-05-30T18:27:57

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

Explanation:

Recall the definition of these functions:

$cot (θ) = \frac{1}{tan (θ)}$ $cot(theta)=1/tan(theta)$
$sec (θ) = \frac{1}{cos (θ)}$ $sec(theta)=1/cos(theta)$
$csc (θ) = \frac{1}{sin (θ)}$ $csc(theta)=1/sin(theta)$

Then we want to prove

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

that is equivalent to

$\frac{1}{tan (θ)} \frac{1}{cos (θ)} = \frac{1}{sin (θ)}$ $1/tan(theta)1/cos(theta)=1/sin(theta)$

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

$\frac{1}{tan (θ)} \frac{1}{cos (θ)} = \frac{1}{sin (θ)}$ $1/tan(theta)1/cos(theta)=1/sin(theta)$

$\frac{cos (θ)}{sin (θ)} \frac{1}{cos (θ)} = \frac{1}{sin (θ)}$ $cos(theta)/sin(theta)1/cos(theta)=1/sin(theta)$

$\frac{1}{sin (θ)} = \frac{1}{sin (θ)}$ $1/sin(theta)=1/sin(theta)$ .

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

1 Answer

Explanation:

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

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Explanation:

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