Solve the following equations :
`2tan^(-1) (sin x) = tan^(-1) (2sec x ,) 0 lt x lt pi/2 `

Answer

To solve the equation \( 2\tan^{-1}(\sin x) = \tan^{-1}(2 \sec x) \) for \( 0 < x < \frac{\pi}{2} \), we will follow these steps:

### Step 1: Use the double angle formula for tangent
We know that:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
Let \( \theta = \tan^{-1}(\sin x) \). Then, we can rewrite the left-hand side:
\[ 2\tan^{-1}(\sin x) = \tan(2\tan^{-1}(\sin x)) = \frac{2\sin x}{1 - \sin^2 x} \]
Using the identity \( 1 - \sin^2 x = \cos^2 x \), we can simplify this to:
\[ \tan(2\tan^{-1}(\sin x)) = \frac{2\sin x}{\cos^2 x} \]

### Step 2: Rewrite the right-hand side
The right-hand side can be rewritten using the definition of secant:
\[ \tan^{-1}(2 \sec x) = \tan^{-1}\left(\frac{2}{\cos x}\right) \]

### Step 3: Set the two sides equal
Now we have:
\[ \frac{2\sin x}{\cos^2 x} = \frac{2}{\cos x} \]

### Step 4: Cross multiply to eliminate the fractions
Cross multiplying gives us:
\[ 2\sin x \cdot \cos x = 2 \cdot \cos^2 x \]

### Step 5: Simplify the equation
Dividing both sides by 2 (assuming \( \sin x \) and \( \cos x \) are not zero in the interval \( (0, \frac{\pi}{2}) \)):
\[ \sin x \cdot \cos x = \cos^2 x \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ \sin x = \cos x \]

### Step 7: Solve for \( x \)
From \( \sin x = \cos x \), we can divide both sides by \( \cos x \) (which is non-zero in the interval):
\[ \tan x = 1 \]
Thus, \( x = \frac{\pi}{4} \).

### Final Solution
The solution to the equation \( 2\tan^{-1}(\sin x) = \tan^{-1}(2 \sec x) \) in the interval \( 0 < x < \frac{\pi}{2} \) is:
\[ x = \frac{\pi}{4} \]

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