Question

If \(sin^{-1}(tan\frac{π}{4})-sin^{-1}(\sqrt{\frac{3}{x}}) = \frac{π}{6}\) then \(x\) is a root of the equation

• A. \(x^2-x-6 = 0\)
• B. \(x^2-x-12 = 0\)
• C. \(x^2+x-12 = 0\)
• D. \(x^2+x-6 = 0\)

Hint:

Evaluate sin^-1(tan(pi/4)) = pi/2, then solve for the second term.

Answer 1

The Correct Option is B

Step 1: Idea:
Since $\sin^{-1}1=\pi/2$, the relation says the second inverse sine equals $\pi/3$.

Step 2: Solve:
Take sine of both sides: $\sqrt{3/x}=\sqrt3/2$. Squaring gives $3/x=3/4$, so $x=4$.

Step 3: Factor Check:
$x^2-x-12=(x-4)(x+3)$, whose roots are 4 and -3. So 4 is a root of option (B).

Final Answer:
Option (B). \[ \boxed{\text{(B) } x^2-x-12=0} \]