If sin\( \theta \) = K find tan\(\theta\), 0° \(\leq\) \(\theta\) \(\leq\) 90°.
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Correct Answer: Option C
Explanation:
\(\sin \theta = \frac{k}{1}\)
\(\implies 1^2 = k^2 + adj^2\)
\(adj = \sqrt{1 - k^2}\)
\(\therefore \tan \theta = \frac{k}{\sqrt{1 - k^2}}\)
\(\sin \theta = \frac{k}{1}\)
\(\implies 1^2 = k^2 + adj^2\)
\(adj = \sqrt{1 - k^2}\)
\(\therefore \tan \theta = \frac{k}{\sqrt{1 - k^2}}\)