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If sin^(-1)(tan pi//4)-sin^(-1)(sqrt(3//...

If `sin^(-1)(tan pi//4)-sin^(-1)(sqrt(3//x))-pi//6=0`, then x is a root of the equation

A

`x^(2)-x-6=0`

B

`x^(2)+x-6=0`

C

`x^(2)-x-12=0`

D

`x^(2)+x-12=0`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the equation \( \sin^{-1}(\tan(\frac{\pi}{4})) - \sin^{-1}\left(\frac{\sqrt{3}}{x}\right) - \frac{\pi}{6} = 0 \), we will follow these steps: ### Step 1: Simplify \( \tan(\frac{\pi}{4}) \) We know that: \[ \tan\left(\frac{\pi}{4}\right) = 1 \] Thus, we can rewrite the equation as: \[ \sin^{-1}(1) - \sin^{-1}\left(\frac{\sqrt{3}}{x}\right) - \frac{\pi}{6} = 0 \] ### Step 2: Evaluate \( \sin^{-1}(1) \) The value of \( \sin^{-1}(1) \) is: \[ \sin^{-1}(1) = \frac{\pi}{2} \] Now, substitute this back into the equation: \[ \frac{\pi}{2} - \sin^{-1}\left(\frac{\sqrt{3}}{x}\right) - \frac{\pi}{6} = 0 \] ### Step 3: Rearrange the equation Rearranging gives: \[ \frac{\pi}{2} - \frac{\pi}{6} = \sin^{-1}\left(\frac{\sqrt{3}}{x}\right) \] ### Step 4: Find a common denominator To simplify \( \frac{\pi}{2} - \frac{\pi}{6} \), we find a common denominator. The least common multiple of 2 and 6 is 6: \[ \frac{\pi}{2} = \frac{3\pi}{6} \] Thus: \[ \frac{3\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \] So now we have: \[ \sin^{-1}\left(\frac{\sqrt{3}}{x}\right) = \frac{\pi}{3} \] ### Step 5: Take the sine of both sides Taking the sine of both sides gives: \[ \frac{\sqrt{3}}{x} = \sin\left(\frac{\pi}{3}\right) \] ### Step 6: Evaluate \( \sin\left(\frac{\pi}{3}\right) \) We know: \[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \] Thus: \[ \frac{\sqrt{3}}{x} = \frac{\sqrt{3}}{2} \] ### Step 7: Solve for \( x \) Cross-multiplying gives: \[ \sqrt{3} \cdot 2 = \sqrt{3} \cdot x \] Dividing both sides by \( \sqrt{3} \) (assuming \( \sqrt{3} \neq 0 \)): \[ 2 = x \] ### Step 8: Conclusion Thus, we find: \[ x = 2 \] ### Step 9: Check for the root of the equation We need to check which equation \( x = 2 \) is a root of. We will substitute \( x = 2 \) into the given options (not provided in the question, but we can assume they are quadratic or polynomial equations).
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