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$$\sqrt{3^2 + 3^2\tan^2\theta}$$

$$ = (3)(3\tan\theta) = 9\tan\theta $$

I've simplified it like this but I'm not sure if that's correct.

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    $\begingroup$ Certainly not correct. Is $\sqrt{a^2 + b^2} = ab$ in general? No. Ask instead: what is $1 + \tan^2\theta$ equal to? $\endgroup$
    – Simon S
    Apr 30, 2015 at 20:05

4 Answers 4

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$$\begin{align} \sqrt{3^2 + 3^2\tan^2\theta} &= \sqrt{9(1+\tan^2 \theta)} \\ & = \sqrt 9 \cdot \sqrt {1 + \tan^2 \theta} \\ & = 3\sqrt{\sec^2\theta} \\ &= 3|\sec \theta| \end{align}$$

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$$\sqrt{3^2+3^2\tan^2 x}$$ $$3 \sqrt{1+\tan ^2x}=3\sqrt{\sec^2 x}=3|\sec x|$$

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Hint $$5x+5y=5(x+y) \text{ distributive property}$$ $$\sqrt{9\cdot 16} = \sqrt{9}\cdot\sqrt{16}\text{ root of a product property}$$

$$\begin{align} opposite^2 + adjacent^2 &= hypotenuse^2\\ \frac{opposite^2 + adjacent^2}{adjacent^2}&=\frac{hypotenuse^2}{adjacent^2}\\ \bigg(\frac{opposite}{adjacent}\bigg)^2 + \bigg(\frac{adjacent}{adjacent}\bigg)^2&=\bigg(\frac{hypotenuse}{adjacent}\bigg)^2\\ \tan^2 x + 1 &= \sec^2 x \end{align}$$

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You have

$$\sqrt{9(1-tan^2(x)} = \sqrt{9 \sec^2(x)} = 3\sec(x).$$

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  • $\begingroup$ It should be 1+tan^2(theta). Also, sqrt(sec^2(theta))=|sec(theta)|, rather than just sec(theta). $\endgroup$
    – Esteemator
    Apr 30, 2015 at 20:40

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