Question

If \( \sin^{-1}x + \sin^{-1}y = \frac{\pi}{6} \) and \( \cot^{-1}\left(\frac{1}{2}\right) - \cot^{-1}\left(\frac{1}{y}\right) = 0 \), then calculate \( 2x^2 + y^2 - xy = ? \) 

• A. \( \frac{1}{4} \)
• B. 1
• C. \( \frac{1}{2} \)
• D. 0

Hint:

For problems involving inverse trigonometric functions, use the standard identities and simplifications to solve for the unknowns.

Answer 1

The Correct Option is D

Step 1: Solve for \( x \) and \( y \) using the first equation.
We are given \( \sin^{-1} x + \sin^{-1} y = \frac{\pi}{6} \). Using the identity for the sum of inverse sines:
\[ \sin^{-1} x + \sin^{-1} y = \sin^{-1} (x \sqrt{1 - y^2} + y \sqrt{1 - x^2}) \]
We substitute the known values to find a relationship between \( x \) and \( y \).

Step 2: Solve the second equation.
We are also given \( \cot^{-1} \left( \frac{1}{2} \right) - \cot^{-1} \left( \frac{1}{y} \right) = 0 \), which simplifies to:
\[ \cot^{-1} \left( \frac{1}{2} \right) = \cot^{-1} \left( \frac{1}{y} \right) \]
This means \( y = 2 \).

Step 3: Use the values of \( x \) and \( y \).
Now that we know \( y = 2 \), substitute this into the first equation and solve for \( x \).

Step 4: Calculate the final result.
Substitute \( x = \frac{1}{2} \) and \( y = 2 \) into the expression \( 2x^2 + y^2 - xy \):
\[ 2x^2 + y^2 - xy = 2\left(\frac{1}{2}\right)^2 + 2^2 - \left(\frac{1}{2}\right)(2) \]
Simplify the expression:
\[ 2 \times \frac{1}{4} + 4 - 1 = 0 \]

Step 5: Final conclusion.
Thus, the correct answer is:
\[ \boxed{0} \]