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Thread: Expressing in terms of X

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    Expressing in terms of X

    Would anybody happen to know how to express this in terms of X?
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    Quote Originally Posted by gearshifter View Post
    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
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    Quote Originally Posted by gearshifter View Post
    Would anybody happen to know how to express this in terms of X?
    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, $\displaystyle sec^2(tan^{-1} 7/x)= sec^2(\theta)= \frac{49+ x^2}{49}$.
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