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) ${cot}^{-1}9+{cosec}^1\frac{\sqrt{41}}{4}=\frac{\pi }{4}$
(b) ${tan}^{-1}\frac{1}{7}+{tan}^{-1}\frac{1}{13}={tan}^{-1}\frac{2}{9}$
(c) ${tan}^{-1}2+{tan}^{-1}3=3\pi /4$

Answer 1

(a) ${cosec}^{-1}x={cot}^{-1}\sqrt{x^2-1}$

L.H.S.$={cot}^{-1}9+{cot}^{-1}\sqrt{\left(41/16\right)-1}$

$={cot}^{-1}9+{cot}^{-1}\left(5/4\right)$

$={tan}^{-1}\frac{1}{9}+{tan}^{-1}\frac{4}{5}$

$={tan}^{-1}\frac{\left(1/9\right)+\left(4/5\right)}{1-\left(1/9\right).\left(4/5\right)}={tan}^{-1}\frac{41}{41}$

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

(b) Do yourself

(c) ${tan}^{-1}2+{tan}^{-1}3=\pi +{tan}^{-1}\frac{2+3}{1-2.3},$

$\because xy=2.3=6>1$

$=\pi +{tan}^{-1}\left\{5/\right(-5\left)\right\}$

$=\pi +{tan}^{-1}(-1)=\pi -{tan}^{-1}1$

$\pi -\pi /4=3\pi /4$