sec(x)+tan(x)=t

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

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- Oct 31st 2013, 06:53 AMAaPaCute trigo problem.
sec(x)+tan(x)=t

tan(x)=?(in form of t) - Oct 31st 2013, 10:28 AMtopsquarkRe: Cute trigo problem.
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):

$\displaystyle sin(x) = t~cos(x) - 1$

Hold on to this equation.

Now square both sides and use $\displaystyle cos^2(x) = 1 - sin^2(x)$ 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:

$\displaystyle cos(x) = \frac{2t}{t^2 + 1}$

Now put this expression for cos(x) into the above sin(x) equation. Then

$\displaystyle tan(x) = \frac{sin(x)}{cos(x)} = \frac{t}{2} - \frac{1}{2t}$

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 PMSorobanRe: Cute trigo problem.
Hello, AaPa!

Quote:

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

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

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

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

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

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

. . . . . . .$\displaystyle \tan x \;=\;\frac{t^2-1}{2t}$

- Oct 31st 2013, 07:10 PMAaPaRe: Cute trigo problem.
yes right. i used the second method.