Question

If $2\cot^{-1}(\frac{4}{3}) = \cos^{-1}(\frac{x}{5})$, then the value of $x$ is equal to

• A. $\frac{3}{25}$
• B. $\frac{7}{25}$
• C. $\frac{3}{5}$
• D. $\frac{7}{5}$
• E. $\frac{5}{7}$

Hint:

Converting all inverse ratios to $\tan^{-1}$ simplifies using the standard double-angle and sum formulas for trigonometric functions.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We need to convert an inverse cotangent function into an inverse cosine function to find \(x\).
Step 2: Key Formula or Approach:
First, convert \(\cot^{-1}\) to \(\tan^{-1}\): \(\cot^{-1}(y) = \tan^{-1}(1/y)\).
Then, use the double angle formula for inverse tangent:
\(2\tan^{-1}(A) = \cos^{-1}\left(\frac{1 - A^2}{1 + A^2}\right)\).
Step 3: Detailed Explanation:
Let's simplify the left side.
\(\cot^{-1}\left(\frac{4}{3}\right) = \tan^{-1}\left(\frac{3}{4}\right)\).
So the equation becomes:
\[ 2\tan^{-1}\left(\frac{3}{4}\right) = \cos^{-1}\left(\frac{x}{5}\right) \] Apply the formula \(2\tan^{-1}(A) = \cos^{-1}\left(\frac{1 - A^2}{1 + A^2}\right)\) with \(A = \frac{3}{4}\):
\[ 2\tan^{-1}\left(\frac{3}{4}\right) = \cos^{-1}\left( \frac{1 - (3/4)^2}{1 + (3/4)^2} \right) \] \[ = \cos^{-1}\left( \frac{1 - 9/16}{1 + 9/16} \right) \] \[ = \cos^{-1}\left( \frac{7/16}{25/16} \right) \] \[ = \cos^{-1}\left(\frac{7}{25}\right) \] Equate this to the right side of the original equation:
\[ \cos^{-1}\left(\frac{7}{25}\right) = \cos^{-1}\left(\frac{x}{5}\right) \] Thus, \(\frac{x}{5} = \frac{7}{25}\).
Solve for \(x\):
\[ x = \frac{7 \times 5}{25} = \frac{35}{25} = \frac{7}{5} \] Step 4: Final Answer:
The value of \(x\) is \(\frac{7}{5}\) (Option D).