Question

Inverse circular functions,Principal values of ${sin}^{-1}x,{cos}^{-1}x,{tan}^{-1}x$.
${tan}^{-1}x+{tan}^{-1}y={tan}^{-1}\frac{x+y}{1-xy}$, $xy<1$
$\pi +{tan}^{-1}\frac{x+y}{1-xy}$, $xy>1$.
Using the principal values, express the following as a single angle :
$3{tan}^{-1}\left(\frac{1}{2}\right)+2{tan}^{-1}\left(\frac{1}{5}\right)+{sin}^{-1}\frac{142}{65\sqrt{5}}$.

Answer 1

${sin}^{-1}\frac{x}{a}={tan}^{-1}\frac{x}{\surd (a^2-x^2)}$

$\therefore$ $T_3={tan}^{-1}\frac{142}{31}$

$2{tan}^{-1}x={tan}^{-1}\frac{2x}{1-x^2}$

$\therefore$ $T_2={tan}^{-1}\frac{5}{12}$

$3{tan}^{-1}x={tan}^{-1}\frac{3x-x^3}{1-3x^2}$

$\therefore$ $T_1={tan}^{-1}\frac{11}{2}$

$\therefore$ $E={tan}^{-1}\frac{11}{2}+{tan}^{-1}\frac{5}{12}+{tan}^{-1}\frac{142}{31}$

$=\pi +{tan}^{-1}\frac{\frac{11}{2}+\frac{5}{12}}{1-\frac{55}{24}}+{tan}^{-1}\frac{142}{31}$

Here $ab>1$ and hence $\pi +{tan}^{-1}\left(\right)$

$=\pi -{tan}^{-1}\frac{142}{31}+{tan}^{-1}\frac{142}{31}=\pi$.