Question

Consider, $\alpha ={\sin }^{-1}\left(\frac{2x}{1+x^2}\right),x\in [-1,1],\beta ={\cos }^{-1}\left(\frac{3\cos y-4\sin y}{10}\right),y\in [0,2\pi ]$ and $\gamma =2{\tan }^{-1}\left(z^2-4z+5\right),z\in R$. 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

Answer: 38

Given $\alpha ={\sin }^{-1}\left(\frac{2x}{1+x^2}\right);x\in [-1,1]$
$\therefore \alpha =2{\tan }^{-1}x\forall x\in [-1,1]$
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
$\gamma =2{\tan }^{-1}\left(z^2-4z+5\right),z\in R$
$\gamma =2{\tan }^{-1}\left((z-2{)}^2+1\right)$
$\therefore$ 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 $\gamma =\frac{\pi }{2}$
$\alpha =2{\tan }^{-1}x=\frac{\pi }{6}px=\tan \frac{\pi }{12}=2-\sqrt{3}$
$b=\frac{\pi }{3}\Rightarrow \cos \beta =\frac{3\cos y-4\sin y}{10}=\frac{1}{2}$
$\therefore \frac{3\cos y}{5}-\frac{4\sin y}{5}=1\Rightarrow \cos (y+\theta )=1\text{ where }\tan \theta =\frac{4}{3}$
$y+\theta =2\pi \Rightarrow y=2\pi -\theta =2\pi -{\tan }^{-1}\frac{4}{3}$
$\gamma =\frac{\pi }{2}\Rightarrow \frac{\pi }{2}=2{\tan }^{-1}\left(z^2-4z+5\right)$
$\therefore z^2-4z+5=1\Rightarrow (z-2{)}^2=0\Rightarrow z=2$
$\text{ Now, }x+\tan y+z=(2-\sqrt{3})+\left(\frac{-4}{3}\right)+2=4-\frac{4}{3}-\sqrt{3}=\frac{8-3\sqrt{3}}{3}=\frac{8-\sqrt{27}}{3}\equiv \frac{a-\sqrt{b}}{c}$
$\therefore a=8,b=27,c=3$
$\Rightarrow a+b+c=38$