# Cute trigo problem.

• Oct 31st 2013, 06:53 AM
AaPa
Cute trigo problem.
sec(x)+tan(x)=t

tan(x)=?(in form of t)
• Oct 31st 2013, 10:28 AM
topsquark
Re: Cute trigo problem.
Quote:

Originally Posted by AaPa
sec(x)+tan(x)=t

tan(x)=?(in form of t)

A quick rundown...As typical with me this is a rather "brute force" approach. I'll leave the details to the reader.

First put the equation into sines and cosines. Then solve for sin(x):
$sin(x) = t~cos(x) - 1$
Hold on to this equation.

Now square both sides and use $cos^2(x) = 1 - sin^2(x)$ to put the equation in terms of only cosine. You can factor out a cos(x) from each side. Then solve for cos(x) getting:
$cos(x) = \frac{2t}{t^2 + 1}$

Now put this expression for cos(x) into the above sin(x) equation. Then
$tan(x) = \frac{sin(x)}{cos(x)} = \frac{t}{2} - \frac{1}{2t}$

Notice that this equation does not have the same domain as the original equation. I'll leave the details of that to you as well.

-Dan
• Oct 31st 2013, 05:10 PM
Soroban
Re: Cute trigo problem.
Hello, AaPa!

Quote:

$\sec x+\tan x \:=\:t$

$\text{Write }\tan x\text{ in terms of }t.$

We have:. $\sec x \;=\;t - \tan x$

Square: . $\sec^2\!x \;=\;t^2 - 2t\tan x + \tan^2\!x$

. . . . $\tan^2\!x+1 \;=\;t^2-2t\tan x + \tan^2\!x$

n . . . . $2t\tan x \;=\;t^2-1$

. . . . . . . $\tan x \;=\;\frac{t^2-1}{2t}$
• Oct 31st 2013, 07:10 PM
AaPa
Re: Cute trigo problem.
yes right. i used the second method.