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 !
 A: $$\tan \alpha =2 \implies \tan 2\alpha = \frac{2\cdot 2}{1-2^2}=-\frac{3}{4} \implies tan(\pi-2\alpha)=\frac{3}{4} \\\implies \cos (\pi-2\alpha)=\frac 35. \blacksquare$$
A: Note
\begin{align}
& \tan(2\tan^{-1}2)-\tan(\pi-\cos^{-1}\frac35)\\
=&\frac{2\tan(\tan^{-1}2)}{1-\tan^2(\tan^{-1}2)}+\tan(\cos^{-1}\frac35)
=\frac{2\cdot2}{1-2^2}+ \frac43=0
\end{align}
which leads to
$$2\tan^{-1}2=\pi-\cos^{-1}\frac35$$
A: 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 :)
A: Let $\cos^{-1}\dfrac35=y,\cos y=?$
Using Principal values,  as $1>\dfrac35>0; 0<y<\dfrac\pi2$
$\tan y=\dfrac{+\sqrt{1-\left(\dfrac35\right)^2}}{\dfrac35}=?, y=\tan^{-1}\dfrac43$
$$\pi-2\tan^{-1}2=2\left(\dfrac\pi2-\tan^{-1}2\right)=2\cot^{-1}2$$
Use Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?
and
Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
A: Hint:
Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
$$2\tan^{-1}2=\pi+\tan^{-1}\dfrac{2\cdot2}{1-2^2}=\pi+\tan^{-1}\left(?\right)$$
Now using Principal values $\tan^{-1}(-y)=-\tan^{-1}y$
Again, if $\tan^{-1}\dfrac43=u,0<u<\dfrac\pi2$  as $\dfrac43>0$
$\cos u=\dfrac1{+\sqrt{1+\tan^2y}}=?$
Can you take it home from here?
