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) ${tan}^{-1}\frac{1}{4}+2{tan}^{-1}\frac{1}{5}+{tan}^{-1}\frac{1}{6}+{tan}^{-1}\frac{1}{x}=\frac{\pi }{4}$

(b) ${tan}^{-1}(x-1)+{tan}^{-1}x+{tan}^{-1}(x+1)={tan}^{-1}3x$

Answer 1

(a) The given question can be written as

$({tan}^{-1}\frac{1}{4}+{tan}^{-1}\frac{1}{6})+2{tan}^{-1}\frac{1}{5}{tan}^{-1}1-{tan}^{-1}\frac{1}{x}$

$\therefore$ ${tan}^{-1}\frac{\left(1/4\right)+\left(1/6\right)}{1-\left(1/4\right).\left(1/6\right)}+{tan}^{-1}\frac{2.\left(1/5\right)}{1-\left(1/25\right)}={tan}^{-1}\frac{1-\left(1/x\right)}{1+\left(1/x\right)}$

or ${tan}^{-1}\frac{10}{23}+{tan}^{-1}\frac{5}{12}={tan}^{-1}\frac{x-1}{x+1}$

or ${tan}^{-1}\frac{\left(10/23\right)+\left(5/12\right)}{1-\left(10/23\right).\left(5/12\right)}={tan}^{-1}\frac{x-1}{x+1}$

or $\frac{235}{226}=\frac{x-1}{x+1}$

Hence $(235-226)x=-235-226$.

(b) ${tan}^{-1}(x-1)+{tan}^{-1}(x+1)={tan}^{-1}3x-{tan}^{-1}x$

or ${tan}^{-1}\frac{(x-1)+(x+1)}{1-(x^2-1)}={tan}^{-1}\frac{3x-x}{1+3x.x}$

or $\frac{2x}{2-x^2}=\frac{2x}{1+3x^2}$

or $x+3x^3=2x-x^3$ or $4x^3-x=0$

or $x(4x^2-1)=0$ $\therefore$ $x=0,\pm \frac{1}{2}$.