Question

Consider, $\alpha=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right), x \in[-1,1], \beta=\cos ^{-1}\left(\frac{3 \cos y-4 \sin y}{10}\right), y \in[0,2 \pi]$ and $\gamma=2 \tan ^{-1}\left(z^2-4 z+5\right), z \in R$. If $\alpha, \beta$ and $\gamma$ are interior angles of a triangle such that $(\beta+\gamma)$ is minimum then $x+\tan y+z=\frac{a-\sqrt{b}}{c}$ where, $a, b, c \in N$. Find the least value of $(a+b+c)$.

Answer 1

Given $\alpha=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right) ; x \in[-1,1]$
$\therefore \alpha=2 \tan ^{-1} x \forall x \in[-1,1]$
Hence, range of $\alpha$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$....(1)
$\beta=\cos ^{-1}\left(\frac{3 \cos y-4 \sin y}{10}\right) ; y \in[0,2 \pi]$
Now, $3 \cos y -4 \sin y \in[-5,5]$
$\therefore$ range of $\beta$ is $\left[\frac{\pi}{3}, \frac{2 \pi}{3}\right]$.....(2)
Also
$\gamma=2 \tan ^{-1}\left(z^2-4 z+5\right), z \in R $
$\gamma=2 \tan ^{-1}\left((z-2)^2+1\right)$
$\therefore$ Range of $\gamma$ is $\left[\frac{\pi}{2}, \pi\right)$....(3)
If $\beta+\gamma$ is minimum, then $\beta=\frac{\pi}{3}$ and $\gamma=\frac{\pi}{2}$ Also, $\alpha, \beta, \gamma$ are angles of a triangle.
Now, $\alpha=\frac{\pi}{6}, \beta=\frac{\pi}{3}$ and $\gamma=\frac{\pi}{2}$
$\alpha=2 \tan ^{-1} x =\frac{\pi}{6} p x =\tan \frac{\pi}{12}=2-\sqrt{3}$
$b =\frac{\pi}{3} \Rightarrow \cos \beta=\frac{3 \cos y -4 \sin y }{10}=\frac{1}{2}$
$\therefore \frac{3 \cos y}{5}-\frac{4 \sin y}{5}=1 \Rightarrow \cos (y+\theta)=1 \text { where } \tan \theta=\frac{4}{3} $
$y+\theta=2 \pi \Rightarrow y=2 \pi-\theta=2 \pi-\tan ^{-1} \frac{4}{3} $
$\gamma=\frac{\pi}{2} \Rightarrow \frac{\pi}{2}=2 \tan ^{-1}\left(z^2-4 z+5\right) $
$\therefore z^2-4 z+5=1 \Rightarrow(z-2)^2=0 \Rightarrow z=2 $
$\text { Now, } x+\tan y+z=(2-\sqrt{3})+\left(\frac{-4}{3}\right)+2=4-\frac{4}{3}-\sqrt{3}=\frac{8-3 \sqrt{3}}{3}=\frac{8-\sqrt{27}}{3} \equiv \frac{a-\sqrt{b}}{c}$
$\therefore a=8, b=27, c=3 $
$\Rightarrow a+b+c=38 $