Question

Inverse circular functions,Principal values of ${sin}^{-1}x,{cos}^{-1}x,{tan}^{-1}x$.
${tan}^{-1}x+{tan}^{-1}y={tan}^{-1}\frac{x+y}{1-xy}$, $xy<1$
$\pi +{tan}^{-1}\frac{x+y}{1-xy}$, $xy>1$.
(a) ${tan}^{-1}\left(tan\frac{3\pi }{4}\right)$
(b) ${tan}^{-1}\left(tan\frac{2\pi }{3}\right)$
(c) $cos[{cos}^{-1}\left(\frac{-\sqrt{3}}{2}\right)=\frac{\pi }{6}]$
(d) If $a\le {tan}^{-1}\left(\frac{1-x}{1+x}\right)\le b$ where $0\le x\le 1$

Answer 1

Let the sides be $a-d,a,a+d$
It is understood that $a>d>0$ and from the figure $\angle C$ is greatest and $\angle A$ is smallest.
By given condition $C=2A$
And hence $B=\pi (A+C)=\pi -3A$
Hence by sine rule we have
$\frac{a+d}{\sin C}=\frac{a-d}{\sin A}=\frac{a}{\sin B}$
or $\frac{a+d}{\sin 2A}=\frac{a-d}{\sin A}=\frac{a}{\sin (\pi -3A)}$
$\therefore 2\cos A=\frac{a+d}{a-d}$ and $\frac{a}{a-d}=\frac{\sin 3A}{\sin A}$
$\therefore \frac{a}{a-d}=3-4{\sin }^{2}A=3-4+{\left(2\cos A\right)}^2$
$\therefore \frac{a}{a-d}=1+{\left(\frac{a+d}{a-d}\right)}^2=\frac{4ad}{{(a-d)}^2}$
$a\ne 0$ $\therefore a-d=4d$ or $a=5d$
$\therefore$ sides are $a-d,aa,a+d$
or $4d,5d,6d$
Hence the required ratio $4:5:6$.