Question

Sol. [Key Concept: Split $\tan^{- 1}\frac{3 - 4tan\ x}{4 + 3tan\ x} = \tan^{- 1}\frac{3}{4} - \tan^{- 1}(tan\ x)$ and differentiate.]

$$y = \tan^{- 1}\left( \frac{\frac{3}{4} - tanx}{1 + \frac{3}{4}tanx} \right) + 2\tan^{- 1}\left( \frac{x}{1 + \sqrt{1 - x^{2}}} \right) $$For $2\tan^{- 1}\left( \frac{x}{1 + \sqrt{1 - x^{2}}} \right)$

put $x = sin\theta$

$$\therefore 2\tan^{- 1}\left( \frac{sin\theta}{1 + cos\theta} \right) = \theta = \sin^{- 1}x $$Hence

$${y = \tan^{- 1}\left( \frac{3}{4} \right) - \tan^{- 1}(tanx) + \sin^{- 1}x }{\frac{dy}{dx} = - 1 + \frac{1}{\sqrt{1 - x^{2}}} }{x = \frac{\sqrt{3}}{2} }{\frac{dy}{dx} = - 1 + 2 = 1 }$$