Question - Trigonometric Identities and Functions
Solution:
To solve the problem, we'll start with the given information:\[ \sec(\theta) = 4 \]\[ \cot(\theta) > 0 \]We want to find the exact values of $$\tan(\theta)$$ and $$\sin(\theta)$$.First, recall the trigonometric identities involving $$\sec(\theta)$$ and $$\tan(\theta)$$:\[ \sec(\theta) = \frac{1}{\cos(\theta)} \]\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \]From $$\sec(\theta) = 4$$, we know that $$\cos(\theta) = \frac{1}{4}$$. Since we are given that $$\cot(\theta) > 0$$, which means $$\cos(\theta)$$ and $$\sin(\theta)$$ must have the same sign, and because $$\cos(\theta) = \frac{1}{4}$$ is positive, $$\sin(\theta)$$ must also be positive. Now we need to find $$\sin(\theta)$$ knowing $$\cos(\theta)$$. Using the Pythagorean identity:\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]Substituting $$\cos(\theta) = \frac{1}{4}$$ into this equation:\[ \sin^2(\theta) + \left(\frac{1}{4}\right)^2 = 1 \]\[ \sin^2(\theta) + \frac{1}{16} = 1 \]\[ \sin^2(\theta) = 1 - \frac{1}{16} \]\[ \sin^2(\theta) = \frac{15}{16} \]Since $$\sin(\theta)$$ is positive (as discussed earlier):\[ \sin(\theta) = \sqrt{\frac{15}{16}} \]\[ \sin(\theta) = \frac{\sqrt{15}}{4} \]Now we can find $$\tan(\theta)$$ using $$\sin(\theta)$$ and $$\cos(\theta)$$:\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \]\[ \tan(\theta) = \frac{\frac{\sqrt{15}}{4}}{\frac{1}{4}} \]\[ \tan(\theta) = \sqrt{15} \]So the exact values are:\[ \tan(\theta) = \sqrt{15} \]\[ \sin(\theta) = \frac{\sqrt{15}}{4} \]
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