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$\theta$ =...

If sin $\theta$ = K find tan $\theta$ , 0° \(...

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.

$\frac{k}{1 - k}$

E.

$\frac{k}{\sqrt{ k^2 - 1}}$

View answer & explanation

Correct answer is C

$\sin \theta = \frac{k}{1}$

$\implies 1^2 = k^2 + adj^2$

$adj = \sqrt{1 - k^2}$

$\therefore \tan \theta = \frac{k}{\sqrt{1 - k^2}}$

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