If `"tan"^(-1)(-x)+"cos"^(-1)((-1)/(2))=pi/2`, then the value of x is equal to

Options

A. `sqrt3`
B. `(-1)/(sqrt3)`
C. `(1)/(sqrt3)`
D. `-sqrt31`

Answer

To solve the equation \( \tan^{-1}(-x) + \cos^{-1}\left(-\frac{1}{2}\right) = \frac{\pi}{2} \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \tan^{-1}(-x) + \cos^{-1}\left(-\frac{1}{2}\right) = \frac{\pi}{2} \]

### Step 2: Find \( \cos^{-1}\left(-\frac{1}{2}\right) \)
From trigonometric values, we know that:
\[ \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \]
Thus, we can replace \( \cos^{-1}\left(-\frac{1}{2}\right) \) with \( \frac{2\pi}{3} \):
\[ \tan^{-1}(-x) + \frac{2\pi}{3} = \frac{\pi}{2} \]

### Step 3: Isolate \( \tan^{-1}(-x) \)
Now, we isolate \( \tan^{-1}(-x) \):
\[ \tan^{-1}(-x) = \frac{\pi}{2} - \frac{2\pi}{3} \]
To simplify the right side, we need a common denominator:
\[ \frac{\pi}{2} = \frac{3\pi}{6}, \quad \frac{2\pi}{3} = \frac{4\pi}{6} \]
Thus,
\[ \tan^{-1}(-x) = \frac{3\pi}{6} - \frac{4\pi}{6} = -\frac{\pi}{6} \]

### Step 4: Use the property of inverse tangent
We know that:
\[ \tan^{-1}(-x) = -\tan^{-1}(x) \]
So we can rewrite the equation:
\[ -\tan^{-1}(x) = -\frac{\pi}{6} \]
This simplifies to:
\[ \tan^{-1}(x) = \frac{\pi}{6} \]

### Step 5: Solve for \( x \)
Taking the tangent of both sides:
\[ x = \tan\left(\frac{\pi}{6}\right) \]
From trigonometric values, we know:
\[ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \]
Thus, the value of \( x \) is:
\[ x = \frac{1}{\sqrt{3}} \]

### Final Answer
The value of \( x \) is \( \frac{1}{\sqrt{3}} \).

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