Question

If \( \sin 48^\circ = p \), then the value of \( \tan 48^\circ \) is:

• A. \( \dfrac{p}{\sqrt{1-p^{2}}} \)
• B. \( \dfrac{\sqrt{1-p^{2}}}{p} \)
• C. \( \dfrac{p}{\sqrt{1+p^{2}}} \)
• D. \( \dfrac{\sqrt{1+p^{2}}}{p} \)

Hint:

For an acute angle, \( \cos \theta = \sqrt{1-\sin^2\theta} \) and \( \tan \theta = \dfrac{\sin\theta}{\cos\theta} \). Always choose the positive square root.

Answer 1

The Correct Option is A

Step 1: Express \(\cos 48^\circ\) in terms of \(p\).
Given \( \sin 48^\circ = p \) and \(48^\circ\) is acute, so \[ \cos 48^\circ = \sqrt{1-\sin^2 48^\circ} = \sqrt{1-p^2}. \]
Step 2: Use \( \tan \theta = \dfrac{\sin \theta}{\cos \theta} \).
\[ \tan 48^\circ = \frac{\sin 48^\circ}{\cos 48^\circ} = \frac{p}{\sqrt{1-p^2}}. \]
Step 3: Conclude.
Hence, \( \tan 48^\circ = \dfrac{p}{\sqrt{1-p^{2}}}. \)