# Expressing in terms of X

• Nov 8th 2008, 08:22 PM
gearshifter
Expressing in terms of X
Would anybody happen to know how to express this in terms of X?
• Nov 8th 2008, 08:32 PM
mr fantastic
Quote:

Originally Posted by gearshifter
Would anybody happen to know how to express this in terms of X?

Read this thread for a way of doing questions like this: http://www.mathhelpforum.com/math-he...xpression.html
• Nov 9th 2008, 01:08 PM
HallsofIvy
Quote:

Originally Posted by gearshifter
Would anybody happen to know how to express this in terms of X?

Saying $tan^{-1}(7/x)= \theta$ is the same as saying $tan(\theta)= 7/x$. Since tangent is "opposite side over near side", if we imagine a right triangle with legs of length 7 and x, $\theta$ is the angle opposite the leg of length 7. Now we can use the Pythagorean theorem to find the length, c, of the hypotenuse: $c^2= a^2+ b^2= 7^2+ x^2= 49+ x^2$ so $c= \sqrt{49+ c^2}$.

Now secant is 1/sine or "hypotenuse divided by opposite side" which, here, is $sec(\theta)= \frac{\sqrt{49+ x^2}{7}$. Finally, $sec^2(tan^{-1} 7/x)= sec^2(\theta)= \frac{49+ x^2}{49}$.