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) ${tan}^{-1}\frac{1}{21}+{tan}^{-1}\frac{1}{13}+{tan}^{-1}(-\frac{1}{8})=0$
(b) ${tan}^{-1}\frac{2}{3}=\frac{1}{2}{tan}^{-1}\frac{12}{5}$
(c) ${tan}^{-1}\frac{1}{4}+{tan}^{-1}\frac{2}{9}=\frac{1}{2}{cos}^{-1}\frac{3}{5}$.

Answer 1

(a) $21=1+5.4$, $13=1+4.3$,
$-1/8=-2/16$ and $6=1+5.3$,
$T_1={tan}^{-1}\frac{5-4}{1+5.4}={tan}^{-1}5-{tan}^{-1}4$
$T_2={tan}^{-1}4-{tan}^{-1}3$
$T_3={tan}^{-1}\frac{-2}{1+15}={tan}^{-1}3-{tan}^{-1}5$
$\therefore$ $sum=0$
Note : You may combine first two terms to get
${tan}^{-1}\left(\frac{34}{272}\right)={tan}^{-1}\left(\frac{1}{8}\right)$
(b) We have to prove ${tan}^{-1}\frac{12}{5}=2{tan}^{-1}\frac{2}{3}$.
Now R.H.S.$={tan}^{-1}\frac{2\left(2/3\right)}{1-\left(4/9\right)}={tan}^{-1}\frac{\left(4/3\right)}{\left(5/9\right)}$.
$={tan}^{-1}\left(12/5\right)=L.H.S.$
(c) L.H.S.$={tan}^{-1}\frac{1}{4}+{tan}^{-1}\frac{2}{9}={tan}^{-1}\frac{1/4+2/9}{1-\left(1/4\right)\left(2/9\right)}$
$={tan}^{-1}\frac{1}{2}=\frac{1}{2}.2{tan}^{-1}\frac{1}{2}$
$=\frac{1}{2}{tan}^{-1}\frac{2\left(1/2\right)}{1-\left(1/4\right)}=\frac{1}{2}{tan}^{-1}\frac{4}{3}=\frac{1}{2}{cos}^{-1}\frac{3}{5}$,