If , then at is equal to: [2026]

Question

If $y = \tan^{- 1} (\frac{3 \cos x - 4 \sin x}{4 \cos x + 3 \sin x}) + 2 \tan^{- 1} (\frac{x}{1 + \sqrt{1 - x^{2}}})$ , then $\frac{d y}{d x}$ at $x = \frac{\sqrt{3}}{2}$ is equal to: [2026]

Smart Achievers · Accepted Answer

(3)

We have, $y = \tan^{- 1} (\frac{3 \cos x - 4 \sin x}{4 \cos x + 3 \sin x}) + 2 \tan^{- 1} (\frac{x}{1 + \sqrt{1 - x^{2}}})$

$= \tan^{- 1} (\frac{\frac{3}{4} - \tan x}{1 + \frac{3}{4} \tan x}) + 2 \tan^{- 1} (\frac{x}{1 + \sqrt{1 - x^{2}}})$

$= \tan^{- 1} (\tan (\frac{3}{4} - x)) + 2 \tan^{- 1} (\frac{x}{1 + \sqrt{1 - x^{2}}})$

$= \frac{3}{4} - x + 2 \tan^{- 1} (\tan \frac{t}{2})$ $[Let x = \sin t, then \frac{x}{1 + \sqrt{1 - x^{2}}} = \tan \frac{t}{2}]$

$= \frac{3}{4} - x + \frac{2 t}{2} = \frac{3}{4} - x + \sin^{- 1} x$

$\frac{d y}{d x} = - 1 + \frac{x}{1 + \sqrt{1 - x^{2}}}$

At $x = \frac{\sqrt{3}}{2},$

$\frac{d y}{d x} |_{\frac{\sqrt{3}}{2}} = - 1 + \frac{1}{\sqrt{1 - \frac{3}{4}}} = - 1 + 2 = 1$

Solution

Q.
If $y = \tan^{- 1} (\frac{3 \cos x - 4 \sin x}{4 \cos x + 3 \sin x}) + 2 \tan^{- 1} (\frac{x}{1 + \sqrt{1 - x^{2}}})$ , then $\frac{d y}{d x}$ at $x = \frac{\sqrt{3}}{2}$ is equal to: [2026]

1 3

2 –1

3 1

4 2

Want Us to Email you About Special Offers & Updates?

Solution

Q. If y=tan–1(3cosx–4sinx4cosx+3sinx)+2tan–1(x1+1–x2), then dydx at x=32 is equal to: [2026]

1 3

2 –1

3 1

4 2

Want Us to Email you About Special Offers & Updates?

Q.
If $y = \tan^{- 1} (\frac{3 \cos x - 4 \sin x}{4 \cos x + 3 \sin x}) + 2 \tan^{- 1} (\frac{x}{1 + \sqrt{1 - x^{2}}})$ , then $\frac{d y}{d x}$ at $x = \frac{\sqrt{3}}{2}$ is equal to: [2026]