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.

Printable View

- Feb 18th 2008, 01:27 PMalgebra2tangent and cotangent
can someone explain this to me:

tan(theta) = 60/x

x = 60/**tan**(theta)

x = 60**cot**(theta)

i dont understand how the tangent switches to cotangent. - Feb 18th 2008, 01:52 PMangel.white
Cotangent is just 1 over tangent so

$\displaystyle cot(x) = \frac 1{tan(x)}$

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

--------------------------------------------------

So your intitial equation is:

$\displaystyle tan(\theta)=\frac{60}x$

then multiply both sides by $\displaystyle \frac{x}{tan(\theta)}$

$\displaystyle \frac{x}{tan(\theta)}*tan(\theta)=\frac{60}x*\frac {x}{tan(\theta)}$

Simplify (cancel out tangents on LHS and x's on RHS) to get your second equation.

$\displaystyle x=\frac{60}{tan(\theta)}$

You can partition the RHS like so:

$\displaystyle x=60*\frac{1}{tan(\theta)}$

And you see that on the RHS 1/tangent = cotangent so

$\displaystyle x=60*cot(\theta)$

Which is your last equation.

--------------------------------------------------

__Some other things to note:__

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

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

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

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