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) ${tan}^{-1}\frac{1}{2}+{tan}^{-1}\frac{1}{3}=\frac{\pi }{4}$.

(b) ${tan}^{-1}\frac{1}{2}+{tan}^{-1}\frac{1}{5}+{tan}^{-1}\frac{1}{8}=\frac{\pi }{4}$

(c) ${tan}^{-1}\frac{3}{4}+{tan}^{-1}\frac{3}{5}-{tan}^{-1}\frac{8}{19}=\frac{\pi }{4}$

Answer 1

(a) ${tan}^{-1}\frac{1/2+1/3}{1-1/2.1/3}$

${tan}^{-1}\frac{5/6}{5/6}={tan}^{-1}1=\frac{\pi }{4}$.

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

$={tan}^{-1}\frac{7}{9}+{tan}^{-1}\frac{1}{8}$

$={tan}^{-1}\frac{7/9+1/8}{1-\left(7/9\right)\left(1/8\right)}$

$={tan}^{-1}\frac{65}{65}={tan}^{-1}1=\frac{\pi }{4}$

c) L.H.S.$={tan}^{-1}\frac{3/4+3/5}{1-\left(3/4\right)\left(3/5\right)}-{tan}^{-1}\frac{8}{19}$

$={tan}^{-1}\frac{27}{11}-{tan}^{-1}\frac{8}{19}$

$={tan}^{-1}\frac{27/11-8/19}{1+\left(27/11\right)\left(8/19\right)}$

$={tan}^{-1}\frac{425/209}{425/209}={tan}^{-1}1=\frac{\pi }{4}$