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