<s>[INST] Prove inverse trigonometric equation $2\tan^{-1}2=\pi-\cos^{-1}\frac{3}{5}$ The question is:

Prove that: $2tan^{-1}2=\pi-cos^{-1}\frac{3}{5}$ Hint: use the fact that $tan(\pi-x)=-tanx$

I did the question without using the Hint, but I don't know how to do it using the hint.
Quick working out of what I've done:
\begin{aligned}
\text { If } \theta &=\tan ^{-1} 2 \\
\tan \theta &=2 \\
0 & < \theta < \frac{\pi}{2}
\end{aligned}
\begin{aligned}
\cos 2 \theta &=2 \cos ^{2} \theta-1 \\
&=2 \times\left(\frac{2}{\sqrt{5}}\right)^{2}-1 \\
&=\frac{3}{5} \\
2 \theta =& \cos ^{-1} \frac{3}{5}, \quad \text { since } 0 < 2\theta < \pi
\end{aligned}
\begin{array}{l}
2 \tan ^{-1} 2=\cos ^{-1} \frac{3}{5} \text { . } \\
\text { Note: } \cos ^{-1} x \text { has point symmetry } \\
\text { in }\left(0, \frac{\pi}{2}\right) \text { . }
\end{array}
$$
\begin{array}{l}
\cos ^{-1} x+\cos ^{-1}(-x)=\pi \\
\cos ^{-1} \frac{3}{5}=\pi-\cos ^{-1}\left(-\frac{3}{5}\right) \\
\therefore \quad 2 \tan ^{-1} 2=\pi-\cos ^{-1}\left(-\frac{3}{5}\right)
\end{array}
$$
But I didn't use the Hint given in the question for this working out. How do I use the hint? Thank you !
 [/INST] It isn't necessary to use the hint. What you have done is correct and valid! Nonetheless, here's the intended method using hint.
Let $\theta=2\tan^{-1}2$ and $\alpha=\pi-\cos^{-1}\frac 3 5$. Then,
$$\tan\theta=\frac{2\tan(\tan^{-1}2)}{1-\tan^2(\tan^{-1}2)}$$
$$\tan\theta=-\frac{4}{3}\quad (*)$$
and using the hint,
$$\tan\alpha=-\tan(\cos^{-1}\frac{3}{5})$$
Using $\cos^{-1}\frac{A}{\sqrt{A^2+B^2}}=\tan^{-1}\frac{B}{A}$, we get
$$\tan\alpha=-\tan(\tan^{-1}\frac{4}{3})$$
$$\tan\alpha=-\frac 4 3\quad (**)$$
From $(*)$ and $(**)$, we conclude
$$\theta=\alpha$$
Or $\text{LHS}=\text{RHS}$ as desired.
Hope this is clear :)
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