Question

solve tan-1 sqrt(x^2 + x) +sin-1 sqrt(x^2 + x + 1) = pi/2

Answer 1

tan-1 [x2+x]1/2 + sin-1 [x2+x+1]1/2 = pi/2

= tan-1[x2+x]1/2 + sin-1 [x2+x+1]1/2 = sin-1 [x2+x+1]1/2 + cos-1 [x2+x+1]1/2

= tan-1 [x2+x]1/2 = cos-1 [x2+x+1]1/2

= cos-1[1/ (x2+x+1)1/2] = cos-1[x2+x+1]1/2

= 1/ (x2+x+1)1/2 = [ x2+x+1]1/2

= x2+x+1 = 1

= x2+x = 0

=x(x+1)  =>  x=0 or x = -1


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Answer 2

Answer:

$x=0 and -1$ are the solution of the given equation.

Step-by-step explanation:

$\begin{aligned}&\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{2} \\&\tan y=\sqrt{x(x+1)} \\&\cos y=\frac{1}{\sqrt{x^{2}+x+1}} \\&\left.y=\cos ^{-1}\right) \frac{1}{\sqrt{x^{2}+x+1}} \\&\cos ^{-1}\left(\frac{1}{\sqrt{x^{2}+x+1}}\right)+\sin ^{-1}\left(\sqrt{x^{2}+x+1}\right)=\frac{\pi}{2}\end{aligned}$

We know that

$\sin ^{-1} y+\cos ^{-1} y=\frac{\pi}{2}$

So $\frac{1}{\sqrt{x^{2}+x+1}}=\sqrt{x^{2}+x+1}$

$\begin{aligned}&x^{2}+x+1=1 \\&x^{2}+x=0 \\&x=0,-1\end{aligned}$

Given equation exists, if

$x^{2}+x \geq 0 and 0 < \sqrt{x^{2}+x+1} \leq 1$

$\left[\because x^{2}+x+1\right.$is always greater than zero

Now, $\quad x^{2}+x \geq 0 and x^{2}+x+1 \leq 1$

$\begin{array}{lllr}\text { or } & x^{2}+x \geq 0 & \text { and } & x^{2}+x \leq 0 \\ \text { or } & x^{2}+x=0 & \text { i.e., } & x(x+1)=0\end{array}$

Hence, $x=0 and -1$ are the solution of the given equation.