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) $tan\left[\frac{1}{2}{cos}^{-1}\left(\frac{\sqrt{5}}{3}\right)\right]$
(b) $tan[2{tan}^{-1}\left(\frac{1}{5}\right)-\frac{\pi }{4}]$

Answer 1

(a) If ${cos}^{-1}\left(\sqrt{5}/3\right)=\theta ,0<\theta <\pi /2$ then
$cos\theta =\sqrt{5}/3$
and $sin\theta =\sqrt{1-{cos}^2\theta }=\sqrt{1-5/9}$
$=\sqrt{4/9}=2/3$
$\therefore$ $tan\left[\frac{1}{2}{cos}^{-1}\frac{\sqrt{5}}{3}\right]=tan\frac{\theta }{2}$
$=\frac{1-cos\theta }{sin\theta }=\frac{[1-(\sqrt{5}/3\left)\right]}{\left(2/3\right)}=\frac{3-\sqrt{5}}{2}$
(b) $\because$ $2{tan}^{-1}\left(\frac{1}{5}\right)={tan}^{-1}\frac{2.\left(1/5\right)}{1-{\left(1/5\right)}^2}={tan}^{-1}\frac{5}{12}$
and $\pi /4={tan}^{-1}1$, we have
$tan\left[2{tan}^{-1}\right(1/5)-\pi /4]$
$=tan\left[{tan}^{-1}\right(5/12)-{tan}^{-1}1]$
$=tan{tan}^{-1}\frac{\left(5/12\right)-1}{1+\left(5/12\right).1}$
$=tan{tan}^{-1}(-\frac{7}{17})=-\frac{7}{17}$