If sin theta K find tan theta 0 leq theta leq 90 - WAEC Mathematics 1994 Question
If sin\( \theta \) = K find tan\(\theta\), 0° \(\leq\) \(\theta\) \(\leq\) 90°.
A
1-K
B
\( \frac{k}{k - 1} \)
C
\( \frac{k}{\sqrt{1 - k^2}} \)
D
\( \frac{k}{1 - k} \)
correct option: c
\(\sin \theta = \frac{k}{1}\)
\(\implies 1^2 = k^2 + adj^2\)
\(adj = \sqrt{1 - k^2}\)
\(\therefore \tan \theta = \frac{k}{\sqrt{1 - k^2}}\)
Please share this, thanks:
Add your answer
No responses