# Thread: Expressing in terms of X

1. ## Expressing in terms of X

Would anybody happen to know how to express this in terms of X?

2. 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

3. 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, $sec^2(tan^{-1} 7/x)= sec^2(\theta)= \frac{49+ x^2}{49}$.