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I need to provide a simplified version of this expression for a homework:

$$ \frac{\cos^{3}x - 2\cdot\cos x + \sec x}{\cos x \cdot \sin^{2}x} $$

Basically, there aren't restrictions. The simpler the final formula, the better. I've been trying to reduce this expression with the main idea of replacing trigonometric function with $\cos$ equivalents (as it appears more frequently, thus could be simplified), but this led me to nowhere.

$$ \frac{\cos^{3}x - 2\cdot\cos x + \frac{1}{\cos x}}{\cos x (1 - \cos^{2}x)} $$

I've also tried other substitutions with no success.

Any tip on direction to take from here are really appreciated. I see there is a lot of work to be done, and I know I haven't walked so much in direction to finishing it. The problem is that I don't have any idea to continue this.

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We can multiply the numerator and denominator by $\cos(x)$ and get, $$\frac{\cos^4(x) - 2 \cdot \cos^2(x) + 1}{\cos^2(x) \cdot \sin^2(x)}$$ $$= \frac{\left(\cos^2(x) - 1 \right)^2}{\cos^2(x) \cdot \sin^2(x)}$$ $$= \frac{\left(-\sin^2(x) \right)^2}{\cos^2(x) \cdot \sin^2(x)}$$ $$= \frac{\sin^4(x)}{\cos^2(x) \cdot \sin^2(x)}$$ $$= \frac{\sin^2(x)}{\cos^2(x)}$$ $$= \tan^2(x)$$

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Try multiplying the numerator (and denominator) by $\cos(x)$. Then the numerator will be a fourth degree expression of the form $t^4-2t^2+1$. (where I have used $t$ in place of $\cos(x)$). This expression is a perfect square and can be factored. You should then be able to cancel some terms from the numerator and denominator.

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  • $\begingroup$ Thank you Joe. I ended up marking the other answer as correct, as it shows the complete solution. I hope you understand! But your explanation was certainly helpful, it was the kind of answer I was waiting. I'm sure I'd continue and find the solution after reading it. Thanks again! Best wishes $\endgroup$ – sidyll May 4 '11 at 1:42

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