Question

Let $2y={\left[{\cot }^{-1}\left(\frac{\sqrt{3}\cos \left(x\right)+\sin \left(x\right)}{\cos \left(x\right)-\sqrt{3}\sin \left(x\right)}\right)\right]}^2$ then $\frac{dy}{dx}=$

• A. $x-\frac{\pi }{6}$
• B. $x+\frac{\pi }{6}$
• C. $2x-\frac{\pi }{6}$
• D. $2x-\frac{\pi }{3}$

Answer 1

The correct option is A

$x-\frac{\pi }{6}$

Determine the derivatives of the given function

Step 1: Simplify the given trigonometric function

Given $2y={\left[{\cot }^{-1}\left(\frac{\sqrt{3}\cos \left(x\right)+\sin \left(x\right)}{\cos \left(x\right)-\sqrt{3}\sin \left(x\right)}\right)\right]}^2$

Divide numerator and denominator by $\cos \left(x\right)$ we get,

$$\begin{gathered} 2y={\left[{\cot }^{-1}\left(\frac{\sqrt{3}+\tan \left(x\right)}{1-\sqrt{3}\tan \left(x\right)}\right)\right]}^2 \\ \Rightarrow 2y={\left[\frac{\pi }{2}-{\tan }^{-1}\left(\frac{\sqrt{3}+\tan \left(x\right)}{1-\sqrt{3}\tan \left(x\right)}\right)\right]}^2 \\ \Rightarrow 2y={\left[\frac{\pi }{2}-{\tan }^{-1}\left(\tan \left(\frac{\pi }{3}+x\right)\right)\right]}^2\left[\because \tan \left(A+B\right)=\frac{\tan \left(A\right)+\tan \left(B\right)}{1-\tan \left(A\right)\tan \left(B\right)}\right] \\ \Rightarrow 2y={\left[\frac{\pi }{2}-\frac{\pi }{3}-x\right]}^2 \\ \Rightarrow 2y={\left[\frac{\pi }{6}-x\right]}^2 \end{gathered}$$

Step 2: Differentiate with respect to $x$ we get,

$$\begin{gathered} 2\frac{dy}{dx}=2\left[\frac{\pi }{6}-x\right]\times \frac{d(-x)}{dx} \\ \Rightarrow \frac{dy}{dx}=-\left[\frac{\pi }{6}-x\right] \\ \Rightarrow \frac{dy}{dx}=x-\frac{\pi }{6} \end{gathered}$$

Therefore the value of $\frac{dy}{dx}=x-\frac{\pi }{6}$

Hence, option A is the correct answer.