Solution of $\tan^{-1} \sqrt{x^2 + x} +\sin^{-1} \sqrt{x^2 + x + 1} = \pi/2$

Question

My attempt at question

$\tan^{-1} \sqrt{x^2 + x} +\sin^{-1} \sqrt{x^2 + x + 1} = \pi/2$

using the identity $\tan^{-1}(\sqrt{x^2 + x} + \sqrt{x^2+x+1/-x^2-x}/[1- \sqrt{x^2+x)(x^2+x+1/-x^2-x)}]$=π/2

$\sqrt{x^2+x} + \sqrt{x^2+x+1/-x^2-x}/[1- \sqrt{(x^2+x)(x^2+x+1/-x^2-x)}$] = 1/0

1- $\sqrt{x^2+x)(x^2+x+1/-x^2-x)}$=0

1=$\sqrt{-x^2-x-1}$

x^2+x+2=0 but this gives complex solution and the answer is 0,-1

Can someone point out my mistake?

Answer 1

You need $\sqrt{x^2+x+1}\le 1$, so $x^2+x\le 0$. On the other hand, you also need $x^2+x\ge0$, so the only points where the left-hand side makes sense are $-1$ and $0$.

Now try substituting $x=0$ and $x=-1$ to find whether they satisfy the equation.

You seem to be using identities that hold, but only when the terms they represent exist. It is true that $$ \arcsin t=\arctan\frac{t}{\sqrt{1-t^2}} $$ when $1-t^2\ne0$, with possible extension when (ab)using $\infty$. For your case it's handier to set $y=x^2+x$, to keep things simpler. Then $$ 1-(\sqrt{y+1})^2=-y^2 $$ and therefore the expression can be written $$ \arctan \sqrt{y}+\arctan\frac{\sqrt{y+1}}{\sqrt{-y}} $$ and the condition $y=0$ clearly shows up.

Going on with further identities is a waste of time.

Answer 2

Let $a=\tan^{-1}\sqrt{x^2+x}$ and $b=\sin^{-1}\sqrt{x^2+x+1}$.

Then $\tan^2 a=x^2+x$ and $\sin^2 b=x^2+x+1$

$\sec^2a=x^2+x+1=\sin^2b=\sin^2(\frac{\pi}{2}-a)=\cos^2a$

So, $\cos^4a=1$ and hence $a=k\pi$, ($k\in\mathbb{Z}$).

$x^2+x=0$.

Answer 3

HInt:

Let $y=\arctan\sqrt{x^2+x}\ge0\implies x^2+x=\tan^2y,0\le y<\dfrac\pi2\implies\cos y\ge0$

$\implies\sin^{-1}(\sec y)=\dfrac\pi2-y$

$\sec y=\sin\left(\dfrac\pi2-y\right)=\cos y$

$\implies\cos y=1$ as $\cos y\ge0$

$\implies\sqrt{x^2+x+1}=1$