Prove that $\tan\bigg[\dfrac{\pi}{4}+\dfrac12 \cos^{-1}\bigg(\dfrac{a}{b}\bigg)\bigg]+\tan\bigg[\dfrac{\pi}{4}-\dfrac12 \cos^{-1}\bigg(\dfrac{a}{b}\bigg)\bigg]=\dfrac{2b}{a}$
I tried substituting $\cos ^{-1}\bigg(\dfrac{a}{b}\bigg)=\theta$ but could not finish it. Please help.
We have: $$\tan(a +b) +\tan(a-b) = \frac{\tan a + \tan b}{1-\tan a \tan b} + \frac{\tan a - \tan b}{1+\tan a \tan b} = \frac{2\sin 2a}{\cos 2a + \cos 2b}\tag 1$$
With $a = \frac{\pi}{4}, b = \frac12 \arccos \left(\frac{a}{b}\right)$, $(1)$ becomes: $$\begin{align} \tan(\frac{\pi}{4}+\frac12 \arccos \left(\frac{a}{b}\right)) + \tan(\frac{\pi}{4}-\frac12 \arccos \left(\frac{a}{b}\right)) \\ = \frac{2\sin \frac{\pi}{2}}{\cos \frac{\pi}{2} + \cos(\arccos(\frac{a}{b}))} = \frac{2b}{a}\end{align}$$
Using $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$$ we have $$\tan\left[\frac{\pi}{4}\pm \frac{1}{2}\cos^{-1}\frac{a}{b}\right]=\frac{1 \pm \tan\frac{1}{2}\cos^{-1}\frac{a}{b}}{1\mp\tan\frac{1}{2}\cos^{-1}\frac{a}{b}}$$ Hence $$\tan\bigg[\dfrac{\pi}{4}+\dfrac12 \cos^{-1}\bigg(\dfrac{a}{b}\bigg)\bigg]+\tan\bigg[\dfrac{\pi}{4}-\dfrac12 \cos^{-1}\bigg(\dfrac{a}{b}\bigg)\bigg]=2\frac{1+\tan^2\left(\frac{1}{2}\cos^{-1}\frac{a}{b}\right)}{1-\tan^2\left(\frac{1}{2}\cos^{-1}\frac{a}{b}\right)}$$
Then by using half angle formula $$\cos\alpha=\frac{1-\tan^2 \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}}$$ we have $$\tan\bigg[\dfrac{\pi}{4}+\dfrac12 \cos^{-1}\bigg(\dfrac{a}{b}\bigg)\bigg]+\tan\bigg[\dfrac{\pi}{4}-\dfrac12 \cos^{-1}\bigg(\dfrac{a}{b}\bigg)\bigg]=\frac{2}{\cos\cos^{-1}\frac{a}{b}}=\frac{2b}{a}$$
To simplify notation, I'll make a slightly different substitution of $\theta = \frac12\cos^{-1}\left(\frac{a}{b}\right)$. Then compute: \begin{align} \tan\left(\frac\pi4 + \theta \right) + \tan\left(\frac\pi4 - \theta \right) &= \frac{\sin\left(\frac\pi4 + \theta\right)}{\cos\left(\frac\pi4 + \theta\right)} + \frac{\sin\left(\frac\pi4 - \theta\right)}{\cos\left(\frac\pi4 - \theta\right)}\\ &= \frac{\sin \frac\pi4\cos\theta + \sin \theta\cos\frac\pi4}{\cos\frac\pi4\cos\theta-\sin\frac\pi4\sin\theta} + \frac{\sin \frac\pi4\cos\theta - \sin \theta\cos\frac\pi4}{\cos\frac\pi4\cos\theta+\sin\frac\pi4\sin\theta}\tag{$*$}\\ &= \frac{\frac{\sqrt2}{2}\cos \theta + \frac{\sqrt2}{2}\sin \theta}{\frac{\sqrt2}{2}\cos \theta - \frac{\sqrt2}{2}\sin\theta} + \frac{\frac{\sqrt2}{2}\cos\theta - \frac{\sqrt2}{2}\sin\theta}{\frac{\sqrt2}{2}\cos\theta + \frac{\sqrt2}{2}\sin\theta}\\ &= \frac{\cos\theta + \sin \theta}{\cos\theta - \sin\theta} + \frac{\cos\theta- \sin\theta}{\cos\theta + \sin \theta}\\ &= \frac{\cos^2\theta + 2\cos\theta\sin\theta + \sin^2\theta + \cos^2\theta - 2\cos\theta\sin\theta + \sin^2\theta}{\cos^2\theta - \sin^2\theta}\\ &= \frac{2}{\cos2\theta}\\ &= \frac{2}{\cos\left(\cos^{-1}\left(\frac ab\right) \right)}\\ &= \frac{2b}{a}. \end{align}
In $(*)$, I've used the angle sum and difference formulas \begin{align} \sin\left(A \pm B\right) &= \sin A\cos B \pm \sin B \cos A,\\ \cos\left(A \pm B\right) &= \cos A \cos B \mp \sin A \sin B. \end{align}
Like Mathistopheles,
let $2x=\arccos\dfrac ab\implies\cos2x=\dfrac ab$
$$F=\tan\left(\dfrac\pi4-x\right)+\tan\left(\dfrac\pi4+x\right)=\dfrac{\sin\left(\dfrac\pi4-x+\dfrac\pi4+x\right)}{\cos\left(\dfrac\pi4-x\right)\cos\left(\dfrac\pi4+x\right)}$$
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$F=\dfrac1{\cos^2\dfrac\pi4-\sin^2x}=\dfrac2{1-2\sin^2x}=\dfrac2{\cos2x}=?$$
