Question

Inverse circular functions,Principal values of ${sin}^{-1}x,{cos}^{-1}x,{tan}^{-1}x$.
${tan}^{-1}x+{tan}^{-1}y={tan}^{-1}\frac{x+y}{1-xy}$, $xy<1$
$\pi +{tan}^{-1}\frac{x+y}{1-xy}$, $xy>1$.
$4{tan}^{-1}\frac{1}{5}-{tan}^{-1}\frac{1}{70}+{tan}^{-1}\frac{1}{99}=\frac{\pi }{4}$

Answer 1

$2{tan}^{-1}\frac{1}{5}={tan}^{-1}\frac{2/5}{1-\left(1/25\right)}={tan}^{-1}\frac{5}{12}$.

$\therefore$ $4{tan}^{-1}\frac{1}{5}=2{tan}^{-1}\frac{5}{12}$

${tan}^{-1}\frac{2\left(5/12\right)}{1-\left(25/144\right)}={tan}^{-1}\frac{120}{119}$.

$\therefore$ $L.H.S.={tan}^{-1}\frac{120}{119}-({tan}^{-1}\frac{1}{70}-{tan}^{-1}\frac{1}{99})$

$={tan}^{-1}\frac{120}{119}-{tan}^{-1}\frac{1/70-1/99}{1+(1+99)\left(1/70\right)}$

$={tan}^{-1}\frac{120}{119}-{tan}^{-1}\frac{29}{6931}$

$={tan}^{-1}\frac{120}{119}-{tan}^{-1}\frac{1}{239}$

$={tan}^{-1}\frac{120/119-1/239}{1+\left(120/119\right).\left(1/239\right)}$

$={tan}^{-1}\frac{120.239-119}{119.239+120}$

$={tan}^{-1}\frac{119.239+(239+119)}{119.239+120}$

$={tan}^{-1}1=\pi /4$

writing $120x=(119+1)x$