Question

The value of tan(cos^-1x) is:

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

Answer 1

The Correct Option is B

To determine the value of \( \tan(\cos^{-1}x) \), we begin by recognizing that if \( \theta = \cos^{-1}x \), then \( \cos \theta = x \). We need to find \( \tan \theta \) given \( \cos \theta = x \).

Recall the identity for tangent in terms of sine and cosine:

\[\tan \theta = \frac{\sin \theta}{\cos \theta}\]

Since \(\cos \theta = x\), we need \(\sin \theta\). Using the Pythagorean identity:

\[\sin^2 \theta + \cos^2 \theta = 1\]

Substitute \(\cos \theta = x\):

\[\sin^2 \theta + x^2 = 1\]

Thus,

\[\sin^2 \theta = 1 - x^2\]

Taking the square root, we get:

\[\sin \theta = \sqrt{1-x^2}\]

Now substitute \(\sin \theta\) and \(\cos \theta\) into the tangent formula:

\[\tan \theta = \frac{\sqrt{1-x^2}}{x}\]

Therefore, the value of \( \tan(\cos^{-1}x) \) is \(\frac{\sqrt{1-x^2}}{x}\), which matches the given option \(\frac{\sqrt{1-x^2}}{x}\).