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 will follow these steps:

### Step 1: Use the double angle formula for tangent
We know that:
\[ 2\tan^{-1}(y) = \tan^{-1\left(\frac{2y}{1-y^2}\right)} \]
Applying this to our equation, we have:
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