Question

If $\sin^{-1}\left(\frac{x}{1+x}\right) = \frac{\pi}{2} - \cos^{-1}\left(\frac{1}{2}\right)$, then $x$ is equal to:

• A. $\frac{1}{2}$
• B. 2
• C. 3
• D. 1
• E. $\frac{1}{4}$

Hint:

Recognizing the $\sin^{-1} \theta + \cos^{-1} \theta = \pi/2$ identity saves having to calculate the actual angles.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We need to solve an equation involving inverse trigonometric functions. The key is to use the complementary angle identity for sine and cosine inverse functions.
Step 2: Key Formula or Approach:
The fundamental identity connecting \( \sin^{-1} \) and \( \cos^{-1} \) is:
\[ \sin^{-1}(A) + \cos^{-1}(A) = \frac{\pi}{2} \] This can be rearranged to:
\[ \sin^{-1}(A) = \frac{\pi}{2} - \cos^{-1}(A) \] Step 3: Detailed Explanation:
The given equation is:
\[ \sin^{-1}\left(\frac{x}{1+x}\right) = \frac{\pi}{2} - \cos^{-1}\left(\frac{1}{2}\right) \] Let's look at the right side of the equation. It is in the form \( \frac{\pi}{2} - \cos^{-1}(A) \) with \( A = \frac{1}{2} \).
Using the identity from Step 2, we can simplify the right side:
\[ \frac{\pi}{2} - \cos^{-1}\left(\frac{1}{2}\right) = \sin^{-1}\left(\frac{1}{2}\right) \] Now, substitute this back into the equation:
\[ \sin^{-1}\left(\frac{x}{1+x}\right) = \sin^{-1}\left(\frac{1}{2}\right) \] If \( \sin^{-1}(u) = \sin^{-1}(v) \), then \( u = v \). Therefore, we can equate the arguments of the \( \sin^{-1} \) functions:
\[ \frac{x}{1+x} = \frac{1}{2} \] Now, we solve this algebraic equation for x. Cross-multiply:
\[ 2(x) = 1(1+x) \] \[ 2x = 1+x \] Subtract x from both sides:
\[ 2x - x = 1 \] \[ x = 1 \] Step 4: Final Answer:
The value of x is 1.