The given function is \(y = \tan^{-1}\left(\sqrt{\frac{1 – \cos x}{1 + \cos x}}\right) \). In this post, we will learn how to find the derivative, \( \frac{dy}{dx} \), of this function with respect to (w.r.t) \( x \).
Solution:
Suppose \( u = \sqrt{\frac{1 – \cos x}{1 + \cos x}} \), they \( y = \tan^{-1}(u) \dots (1)\)
Now simplify the value of \( u \):
\[ \displaystyle u = \sqrt{ \frac{1 – \cos x}{1 + \cos x} } \\ \displaystyle \Rightarrow u = \sqrt{ \frac{2 \sin^2\left(\frac{x}{2}\right)}{2 \cos^2\left(\frac{x}{2}\right)} } \small \text{, using trigonometric identities} \\ \displaystyle \Rightarrow u = \sqrt{ \tan^2\left(\frac{x}{2}\right) } \\ \displaystyle \Rightarrow u = \tan\left(\frac{x}{2}\right) \]Now put the value of \( u \) in (1), so
\[
y = \tan^{-1}\left(\tan\left(\frac{x}{2}\right)\right)
\]
Since \( \tan^{-1}(\tan \alpha) = \alpha \) for \( \alpha \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), we have:
\[ y = \frac{x}{2} \]
Differentiating \( y = \frac{x}{2} \) with respect to \( x \):
\[
\frac{dy}{dx} = \frac{1}{2}
\]
Therefore, \( \frac{dy}{dx} \) for the given function is:
\[
\fbox{$ \frac{d}{dx} \left[ \tan^{-1}\left(\sqrt{\frac{1 – \cos x}{1 + \cos x}}\right) \right] = \frac{1}{2}$}
\]
Please let me know in the comments if you find any error in this solution.