Question 56

If $$\tan \theta = \frac{2}{3}, then \frac{3 \sin \theta - 4 \cos \theta}{3 \sin \theta + 4 \cos \theta}$$ is equal to:

Solution

$$\tan \theta = \frac{2}{3}$$

$$\frac{prependicular}{base}=\frac{2}{3},hypotenuse=\sqrt{\ 13}$$

$$\sin=\frac{p}{h},\cos=\frac{b}{h}$$

$$ \frac{3 \sin \theta - 4 \cos \theta}{3 \sin \theta + 4 \cos \theta}$$

$$\frac{\left(\frac{6}{\sqrt{\ 3}}-\frac{12}{\sqrt{\ 3}}\right)}{\frac{6}{\sqrt{\ 3}}+\frac{12}{\sqrt{\ 3}}}=-\frac{6}{18}=-\frac{1}{3}$$

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