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

Printable View

- Nov 8th 2008, 08:22 PMgearshifterExpressing in terms of X
Would anybody happen to know how to express this in terms of X?

- Nov 8th 2008, 08:32 PMmr fantastic
Read this thread for a way of doing questions like this: http://www.mathhelpforum.com/math-he...xpression.html

- Nov 9th 2008, 01:08 PMHallsofIvy
Saying $\displaystyle tan^{-1}(7/x)= \theta$ is the same as saying $\displaystyle tan(\theta)= 7/x$. Since tangent is "opposite side over near side", if we imagine a right triangle with legs of length 7 and x, $\displaystyle \theta$ is the angle opposite the leg of length 7. Now we can use the Pythagorean theorem to find the length, c, of the hypotenuse: $\displaystyle c^2= a^2+ b^2= 7^2+ x^2= 49+ x^2$ so $\displaystyle c= \sqrt{49+ c^2}$.

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