Solve for `x :` `2tan^(-1)(sinx)=tan^(-1)(2secx),\ x!=pi/2`

Explanation

To solve the equation \(2\tan^{-1}(\sin x) = \tan^{-1}(2\sec x)\), we can follow these steps:

### Step 1: Use the addition formula for inverse tangent
We know that:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \quad \text{if } ab < 1 \]
Thus, we can express \(2\tan^{-1}(\sin x)\) as:
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