If y = tan^-1 ( 3cos x - 4sin x 4cos x + 3sin x ) + 2tan^-1 ( x 1+√1-x^2 ), then dy dx at x = √3 2 is equal to:

Let $y = y_1 + y_2$y=y1​+y2​

$$y_1 = \tan^{-1}\left(\dfrac{3\cos x - 4\sin x}{4\cos x + 3\sin x}\right)$$y1​=tan−1(4cosx+3sinx3cosx−4sinx​)

Dividing the numerator and denominator by $4\cos x$4cosx:

$$y_1 = \tan^{-1}\left(\dfrac{\dfrac{3}{4} - \tan x}{1 + \dfrac{3}{4}\tan x}\right)$$y1​=tan−1​1+43​tanx43​−tanx​​

Using the formula $\tan^{-1}\left(\dfrac{A - B}{1 + AB}\right) = \tan^{-1}A - \tan^{-1}B$tan−1(1+ABA−B​)=tan−1A−tan−1B:

$$y_1 = \tan^{-1}\left(\dfrac{3}{4}\right) - \tan^{-1}(\tan x) = \tan^{-1}\left(\dfrac{3}{4}\right) - x$$y1​=tan−1(43​)−tan−1(tanx)=tan−1(43​)−x

Differentiating with respect to $x$x:

$$\dfrac{dy_1}{dx} = -1$$dxdy1​​=−1

Now, $y_2 = 2\tan^{-1}\left(\dfrac{x}{1+\sqrt{1-x^2}}\right)$y2​=2tan−1(1+1−x2​x​)

Substituting $x = \sin \theta$x=sinθ:

$$y_2 = 2\tan^{-1}\left(\dfrac{\sin \theta}{1+\cos \theta}\right)$$y2​=2tan−1(1+cosθsinθ​)

$$y_2 = 2\tan^{-1}\left(\dfrac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)}\right)$$y2​=2tan−1(2cos2(θ/2)2sin(θ/2)cos(θ/2)​)

$$y_2 = 2\tan^{-1}\left(\tan\dfrac{\theta}{2}\right) = 2\left(\dfrac{\theta}{2}\right) = \theta = \sin^{-1}x$$y2​=2tan−1(tan2θ​)=2(2θ​)=θ=sin−1x

Differentiating with respect to $x$x:

$$\dfrac{dy_2}{dx} = \dfrac{1}{\sqrt{1-x^2}}$$dxdy2​​=1−x2​1​

At $x = \dfrac{\sqrt{3}}{2}$x=23​​:

$$\dfrac{dy_2}{dx} = \dfrac{1}{\sqrt{1 - \left(\dfrac{\sqrt{3}}{2}\right)^2}} = \dfrac{1}{\sqrt{1 - \dfrac{3}{4}}} = \dfrac{1}{\sqrt{\dfrac{1}{4}}} = 2$$dxdy2​​=1−(23​​)2​1​=1−43​​1​=41​​1​=2

Therefore, $\dfrac{dy}{dx} = \dfrac{dy_1}{dx} + \dfrac{dy_2}{dx} = -1 + 2 = 1$dxdy​=dxdy1​​+dxdy2​​=−1+2=1

Answer: $1$1