Question

Inverse circular functions,Principal values of ${sin}^{-1}x,{cos}^{-1}x,{tan}^{-1}x$.

${tan}^{-1}x+{tan}^{-1}y={tan}^{-1}\frac{x+y}{1-xy}$, $xy<1$

$\pi +{tan}^{-1}\frac{x+y}{1-xy}$, $xy>1$.

$2{tan}^{-1}\left(\sqrt{\frac{a-b}{a+b}}tan\frac{\theta }{2}\right)={cos}^{-1}\frac{acos\theta +b}{a+bcos\theta }$.

Answer 1

we have that $2{tan}^{-1}x={cos}^{-1}\frac{1-x^2}{1+x^2}$.

Choose $x=\sqrt{\left(\frac{a-b}{a+b}\right)}tan\frac{\theta }{2}$

$\therefore$ $2{tan}^{-1}\left[\sqrt{\left(\frac{a-b}{a+b}\right)}tan\frac{\theta }{2}\right]$

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

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

$={cos}^{-1}\frac{+b\left[{cos}^2\right(\theta /2)+{sin}^2(\theta /2\left)\right]}{a\left[{cos}^2\right(\theta /2)+{sin}^2(\theta /2\left)\right]}$

$+b\left[{cos}^2\right(\theta /2)-{sin}^2(\theta /2\left)\right]$

$={cos}^{-1}\frac{acos\theta +b}{a+bcos\theta }$.