Question

Inverse circular functions,Principal values of ${sin}^{-1}x,{cos}^{-1}x,{tan}^{-1}x$.
${tan}^{-1}x+{tan}^{-1}y={tan}^{-1}\frac{x+y}{1-xy}$, $xy<1$
$\pi +{tan}^{-1}\frac{x+y}{1-xy}$, $xy>1$.
Prove that
(a) $2{tan}^{-1}\frac{1}{5}+{sec}^{-1}\frac{5\sqrt{2}}{7}+2{tan}^{-1}\frac{1}{8}=\frac{\pi }{4}$
(b) ${cos}^{-1}\frac{12}{13}+2{cos}^{-1}\sqrt{\left(\frac{64}{65}\right)}+{cos}^{-1}\sqrt{\left(\frac{49}{50}\right)}={cos}^{-1}\frac{1}{\sqrt{2}}$

Answer 1

(a) L.H.S.

$=2({tan}^{-1}\frac{1}{5}+{tan}^{-1}\frac{1}{8})+{tan}^{-1}\sqrt{[{\left(\frac{5\sqrt{2}}{7}\right)}^2-1]}$

$=2{tan}^{-1}\frac{1/5+1/8}{1-\left(1/5\right).\left(1/8\right)}+{tan}^{-1}\frac{1}{7}$

$\because {sec}^{-1}x={tan}^{-1}\sqrt{x^2-1}$

$=2{tan}^{-1}\frac{13}{39}+{tan}^{-1}\frac{1}{7}=2{tan}^{-1}\frac{1}{3}+{tan}^{-1}\frac{1}{7}$

$={tan}^{-1}\frac{2.1/3}{1-1/9}+{tan}^{-1}\frac{1}{7}={tan}^{-1}\frac{3}{4}+{tan}^{-1}\frac{1}{7}$

$={tan}^{-1}\frac{3/4+1/7}{1-\left(3/4\right).\left(1/7\right)}={tan}^{-1}\frac{25}{25}$

$={tan}^{-1}1=\frac{\pi }{4}$.

(b) ${cos}^{-1}x={tan}^{-1}\frac{\sqrt{1-x^2}}{x}$.

$\therefore$ ${cos}^{-1}\frac{12}{13}={tan}^{-1}\frac{\sqrt{1-144/169}}{12/13}={tan}^{-1}\frac{5}{12}$

${cos}^{-1}\sqrt{\left(\frac{49}{50}\right)}={tan}^{-1}\frac{\sqrt{1-49/50}}{\surd \left(49/50\right)}={tan}^{-1}\frac{1}{7}$.

Similarly ${cos}^{-1}\sqrt{64/65}={tan}^{-1}\left(1/8\right)$,

${cos}^{-1}\left(1/\sqrt{2}\right)={tan}^{-1}1=\pi /4$

Thus we have to prove that

${tan}^{-1}\left(5/12\right)+2{tan}^{-1}\left(1/8\right)+{tan}^{-1}\left(1/7\right)={tan}^{-1}1$

or ${tan}^{-1}\frac{5/12+1/7}{1-\left(5/12\right).\left(1/7\right)}={tan}^{-1}1-{tan}^{-1}\frac{2.1/8}{1-1/64}$

or ${tan}^{-1}\left(47/49\right)={tan}^{-1}1-{tan}^{-1}\left(16/63\right)$

or ${tan}^{-1}\frac{47}{79}={tan}^{-1}\frac{1-\left(16/63\right)}{1+\left(16/63\right)}={tan}^{-1}\frac{47}{79}$.