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

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

Answer 1

Let ${cos}^{-1}\left(a/b\right)=\theta$ so that $cos\theta =1/b$.

$\therefore$ L.H.S.$=tan(\pi /4+\theta /2)+tan(\pi /4-\theta /2)$

$=\frac{1+tan\left(\theta /2\right)}{1-tan\left(\theta /2\right)}+\frac{1-tan\left(\theta /2\right)}{1+tan\left(\theta /2\right)}$

$=\frac{{[1+tan(\theta /2\left)\right]}^2+{[1-tan(\theta /2\left)\right]}^2}{1-{tan}^2\left(\theta /2\right)}$

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

$=2\frac{{cos}^2\left(\theta /2\right)+{sin}^2\left(\theta /2\right)}{{cos}^2\left(\theta /2\right)-{sin}^2\left(\theta /2\right)}=2.\frac{1}{cos\theta }$

$=2.\frac{1}{a/b}=\frac{2b}{a}$.