Find $x$ if $\cot^{-1}\left(\frac{1}{x}\right)+\cos ^{-1}(-x)+\tan^{-1}(x)=\pi$

Question

if $x \lt 0$ Then Find value of $$\frac{(1-x^2)^{\frac{3}{2}}}{x^2}$$ if

$$ \cot^{-1} \left(\frac{1}{x}\right)+\cos^{-1}(-x)+\tan^{-1}(x)=\pi$$

My try:

Since $x \lt 0$ we have $$ \cot^{-1}\left(\frac{1}{x}\right)=\pi +\tan ^{-1}x$$

Also $$\cos^{-1}(-x)=\pi -\cos ^{-1}x$$

Hence the given equation becomes

$$\pi +2 \tan^{-1} x+\pi -\cos^{-1}{x}=\pi$$

$$\pi+2 \tan^{-1}x=\cos^{-1}x$$

taking $\cos$ both sides we get

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

Now how to proceed further?

Answer 1

$$\cot^{-1}\left(\frac{1}{x}\right)+\cos ^{-1}(-x)+\tan^{-1}(x)=\pi$$

Suppose $\tan^{-1}(x) = y \implies x = \tan y \implies \cot^{-1}(1/x) = y$

Let $\cos ^{-1}(-x) = z$

$ 2y + z = \pi \implies z = \pi - 2y$

Taking tangents, $\tan(\pi - 2y) = \tan z \implies -\tan(2y) = \sin z/\cos z $

We get $-\tan 2y = \frac{\sqrt{1 - (-x)^2}}{-x}$ using $\cos z = -x$

$\frac{2 \tan y}{1 - \tan^2(y)} = \frac{\sqrt{1 - (-x)^2}}{x}$

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

$\implies \frac{(1-x^2)\sqrt{(1-x^2)}}{x^2} = 2$

Answer 2

A further step to a solution is along the following lines: Note that if $$cot^{-1}\left(\frac{1}{x}\right)=\theta \Rightarrow{cot(\theta)=\frac{1}{x}=\frac{1}{tan(\theta)}}\Rightarrow{x=tan(\theta)}$$ Therefore $$cot^{-1}\left(\frac{1}{x}\right)=tan^{-1}\left(x\right)$$ The original problem becomes $$2tan^{-1}\left(x\right)+cos^{-1}\left(-x\right)=\pi$$ Taking the $cosine$ of both sides gives: $$cos\left[2tan^{-1}\left(x\right)+cos^{-1}\left(-x\right)\right]=-1$$ Using the formula $cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$: $$cos\left[2tan^{-1}\left(x\right)\right]cos\left[cos^{-1}\left(-x\right)\right]-sin\left[2tan^{-1}\left(x\right)\right]sin\left[cos^{-1}\left(-x\right)\right]=-1 \tag{1}$$ Now, recall from Calculus II that if we let $\theta=tan^{-1}\left(x\right)\Rightarrow {x=tan(\theta)}$, which means (by using a right angle triangle) that: $$sin(\theta)=\frac{x}{\sqrt[]{1+x^2}}$$ and $$cos(\theta)=\frac{1}{\sqrt[]{1+x^2}}$$ Therefore $$cos\left[2tan^{-1}\left(x\right)\right]=cos\left[2\theta\right]=2cos^2{\theta}-1=2\left(\frac{1}{\sqrt[]{1+x^2}}\right)^2-1 \tag{2}$$

$$cos\left[cos^{-1}\left(-x\right)\right]=-x \tag{3}$$

$$sin\left[2tan^{-1}\left(x\right)\right]=sin\left[2\theta\right]=2sin(\theta)cos(\theta)=2\left({\frac{x}{\sqrt[]{1+x^2}}}\right)\left({\frac{1}{\sqrt[]{1+x^2}}}\right) \tag{4}$$

$$sin\left[cos^{-1}\left(-x\right)\right]=\sqrt[]{1-x^2} \tag{5}$$

Substituting equations $(2), (3)$ and $(4)$ and $(5)$ into $(1)$ gives: $$\left[2\left(\frac{1}{\sqrt[]{1+x^2}}\right)^2-1\right]{\left(-x\right)}-2\left({\frac{x}{\sqrt[]{1+x^2}}}\right)\left({\frac{1}{\sqrt[]{1+x^2}}}\right)\sqrt[]{1-x^2}=-1$$

Simplifying and collecting like terms gives: $$x^3 + x^2 - x + 1 -2x\sqrt[]{1-x^2}=0$$ Now, to try and solve for $x$, first let: $$x^3 + x^2 - x + 1=2x\sqrt[]{1-x^2}$$ $$\Rightarrow {\left(x^3 + x^2 - x^2 + 1\right)^2=\left(2x\sqrt[]{1-x^2}\right)^2}$$ Again, simplifying and collecting like terms, gives: $$x^6+2x^5-x^4+3x^2-2x+1=0$$ But (after much work!) $$x^6+2x^5-x^4+3x^2-2x+1=\left(x^3+x^2+x-1\right)^2=0$$

Finally, the value of $x$ can be found by solving the cubic equation: $$x^3+x^2+x-1=0 \tag{6}$$ When $x=0$, the equation is negative, and when $x=1$ the equations is positive. Hence, since the equation is a polynomial, by the Intermediate Value Theorem (IVT) the equation $(5)$ has a real root between $0$ and $1$. Unfortunately, you may have to use the (very complicated) formula for finding the roots of a cubic equation or a numerical solver to determine an accurate value of $x$. Using a numerical solver at this point kind of nullifies all the previous work, since you could have just used a numerical solver for the original problem.