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) Solve the equation
${tan}^{-1}2x+{tan}^{-1}3x=n\pi +\left(\pi /4\right)$.
(b) Find all the positive integral solutions of
${tan}^{-1}x+{cos}^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right)={sin}^{-1}\left(\frac{3}{\sqrt{10}}\right)$

Answer 1

(a) L.H.S.$={tan}^{-1}\frac{2x+3x}{1-2x.3x}=n\pi +\frac{3\pi }{4}$

or $\frac{5x}{1-6x^2}tan\frac{3\pi }{4}=-1$

or $6x^2-5x-1=0$

or $(x-1)(6x-1)=0$ $\therefore$ $x=1$, $-1/6$

(b) The given equation can be written as

${tan}^{-1}x+{tan}^{-1}\frac{1}{y}={tan}^{-1}3$

or ${tan}^{-1}\frac{x+1/y}{1-x.1/y}={tan}^{-1}3$

$\therefore$ $xy+1=3(y-x)$. $\therefore$ $y=\frac{3x+1}{3-x}$

There is only one equation in two variables, so we choose $x=k$

$\therefore$ $y=\frac{3k+1}{3-k}$. Further, we have to find only $+ive$ and integral solutions. This is possible only when $k=1$ or $2$ and for no other value.

Hence $(1,2),(2,7)$ are the only two $+ive$ integral solutions.