Solveeit Logo
Login

Question

Question: Let $f(x) = (sin(tan^{-1} x) + sin(cot^{-1} x))^2 -1, |x|>1$. If $\frac{dy}{dx} = \frac{1}{2} \frac{...

Let f(x)=(sin(tan−1x)+sin(cot−1x))2−1,∣x∣>1f(x) = (sin(tan^{-1} x) + sin(cot^{-1} x))^2 -1, |x|>1. If dydx=12ddx(sin−1(f(x)))\frac{dy}{dx} = \frac{1}{2} \frac{d}{dx}(sin^{-1} (f(x))) and y(3)=π6y(\sqrt{3}) = \frac{\pi}{6}, then y(−3)y(-\sqrt{3}) is equal to

A

π3\frac{\pi}{3}

B

2π3\frac{2\pi}{3}

C

−π6-\frac{\pi}{6}

D

5π6\frac{5\pi}{6}

Answer

5π6\frac{5\pi}{6}

Explanation

Solution

First, simplify f(x)=(sin⁡(tan⁡−1x)+sin⁡(cot⁡−1x))2−1f(x) = (\sin(\tan^{-1} x) + \sin(\cot^{-1} x))^2 -1 to f(x)=2xx2+1f(x) = \frac{2x}{x^2+1}.

Express sin⁡−1(f(x))\sin^{-1}(f(x)) in terms of tan⁡−1x\tan^{-1} x for ∣x∣>1|x|>1:

  • If x>1x>1, sin⁡−1(f(x))=π−2tan⁡−1x\sin^{-1}(f(x)) = \pi - 2\tan^{-1} x.
  • If x<−1x<-1, sin⁡−1(f(x))=−π−2tan⁡−1x\sin^{-1}(f(x)) = -\pi - 2\tan^{-1} x.

Differentiate sin⁡−1(f(x))\sin^{-1}(f(x)) to get −21+x2-\frac{2}{1+x^2} for both cases.

Calculate dydx=12(−21+x2)=−11+x2\frac{dy}{dx} = \frac{1}{2} \left(-\frac{2}{1+x^2}\right) = -\frac{1}{1+x^2}.

Integrate to find y(x)=−tan⁡−1x+Cy(x) = -\tan^{-1} x + C. Assume CC is constant across the domain.

Use y(3)=π6y(\sqrt{3}) = \frac{\pi}{6} to find C=π2C = \frac{\pi}{2}.

Calculate y(−3)=−tan⁡−1(−3)+π2=π3+π2=5π6y(-\sqrt{3}) = -\tan^{-1}(-\sqrt{3}) + \frac{\pi}{2} = \frac{\pi}{3} + \frac{\pi}{2} = \frac{5\pi}{6}.