# tangent and cotangent

• Feb 18th 2008, 01:27 PM
algebra2
tangent and cotangent
can someone explain this to me:

tan(theta) = 60/x

x = 60/tan(theta)

x = 60cot(theta)

i dont understand how the tangent switches to cotangent.
• Feb 18th 2008, 01:52 PM
angel.white
Quote:

Originally Posted by algebra2
can someone explain this to me:

tan(theta) = 60/x

x = 60/tan(theta)

x = 60cot(theta)

i dont understand how the tangent switches to cotangent.

Cotangent is just 1 over tangent so
$cot(x) = \frac 1{tan(x)}$

In the same way that cosecant is 1 over sine, and secant is 1 over cosine.

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So your intitial equation is:
$tan(\theta)=\frac{60}x$

then multiply both sides by $\frac{x}{tan(\theta)}$
$\frac{x}{tan(\theta)}*tan(\theta)=\frac{60}x*\frac {x}{tan(\theta)}$

Simplify (cancel out tangents on LHS and x's on RHS) to get your second equation.
$x=\frac{60}{tan(\theta)}$

You can partition the RHS like so:
$x=60*\frac{1}{tan(\theta)}$

And you see that on the RHS 1/tangent = cotangent so
$x=60*cot(\theta)$

Which is your last equation.

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Some other things to note:

since $cot(\theta) = \frac 1{tan(\theta)}$ you can multiply both sides by tangent over cotangent and get $tan(\theta)=\frac 1{cot(\theta)}$

and also since $tan(\theta) = \frac{sin(\theta)}{cos(\theta)}$

it follows that $cot(\theta) = \frac 1{tan(\theta)} = \frac{1}{\frac{sin(\theta)}{cos(\theta)}} = \frac{cos(\theta)}{sin(\theta)}$

so $cot(\theta)=\frac{cos(\theta)}{sin(\theta)}$