If tan ^{-1} sqrt{x^2+x}+sin ^{-1} sqrt{x^2+x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the equation: $	an ^{-1} \sqrt{x^2+x}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$ Since $\sin ^{-1} 	heta + \cos ^{-1} 	heta = \frac{\pi}{2}

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Given the equation: $	an ^{-1} \sqrt{x^2+x}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$ Since $\sin ^{-1} 	heta + \cos ^{-1} 	heta = \frac{\pi}{2}$,we can write: $	an ^{-1} \sqrt{x^2+x} = \frac{\pi}{2} - \sin ^{-1} \sqrt{x^2+x+1} = \cos ^{-1} \sqrt{x^2+x+1}$ Let $	an ^{-1} \sqrt{x^2+x} = 	heta$. Then $	an 	heta = \sqrt{x^2+x}$. This implies $\cos 	heta = \frac{1}{\sqrt{1 + (x^2+x)}}$. Thus,$\cos ^{-1} \left( \frac{1}{\sqrt{x^2+x+1}} ight) = \cos ^{-1} \sqrt{x^2+x+1}$. Comparing the arguments: $\frac{1}{\sqrt{x^2+x+1}} = \sqrt{x^2+x+1}$ $1 = x^2+x+1$ $x^2+x = 0$ $x(x+1) = 0$ So,$x = 0$ or $x = -1$. Checking the domain of $\sin ^{-1} \sqrt{x^2+x+1}$,we require $0 \le x^2+x+1 \le 1$,which implies $x^2+x \le 0$. For $x=0$,$x^2+x=0$ (valid). For $x=-1$,$x^2+x=0$ (valid). However,$	an ^{-1} \sqrt{x^2+x}$ requires $x^2+x \ge 0$. Thus,$x^2+x=0$ is the only solution,giving $x=0$ or $x=-1$.

If $\tan ^{-1} \sqrt{x^2+x}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$,then the value of $x$ is

Similar Questions

If $\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\frac{\pi}{2}$, then $1-x y-y z-z x$ is equal to

If $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{4}$,then $\cos ^{-1}\left(\frac{12}{13} \cos x+\frac{5}{13} \sin x\right)$ is equal to

If $\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{3}{5}+\sin ^{-1} x=\frac{\pi}{2}$,then $x=$

If $3 \sin^{-1}(\frac{2x}{1+x^2}) - 4 \cos^{-1}(\frac{1-x^2}{1+x^2}) + 2 \tan^{-1}(\frac{2x}{1-x^2}) = \frac{\pi}{3}$, then find the value of $x$.

The value of $\cot ^{-1}\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right]$,where $x \in\left(0, \frac{\pi}{4}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module