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) Given $0\le x\le \frac{1}{2}$ then the value of

$tan[{sin}^{-1}\{\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^2}}{\sqrt{2}}\}-{sin}^{-1}x]$ is

(b) If $\alpha ={sin}^{-1}\frac{4}{5}+{sin}^{-1}\frac{1}{3}$

and $\beta ={cos}^{-1}\frac{4}{5}+{cos}^{-1}\frac{1}{3}$,

Answer 1

(a) . Put $x=sin\theta$

${sin}^{-1}(\frac{1}{\sqrt{2}}sin\theta +\frac{1}{\sqrt{2}}cos\theta )={sin}^{-1}sin(\theta +\frac{\pi }{4})=\theta +\frac{\pi }{4}$

$\therefore$ $E=tan[\theta +\frac{\pi }{4}-\theta ]=tan\frac{\pi }{4}=1$

(b) .

$\alpha ={sin}^{-1}[\frac{4}{5}\sqrt{1-\frac{1}{9}}+\frac{1}{3}\sqrt{1-\frac{16}{25}}]$

$={sin}^{-1}[\frac{8\sqrt{2}}{15}+\frac{3}{15}]={sin}^{-1}\left(\frac{8\sqrt{2}+3}{15}\right)$

Since $\frac{8\sqrt{2}+3}{15}<1$ $\therefore \alpha <\frac{\pi }{2}$............(I)

$\beta =(\frac{\pi }{2}-{sin}^{-1}\frac{4}{5}+\frac{\pi }{2}-{sin}^{-1}\frac{1}{3})=\pi -\alpha >\frac{\pi }{2}$ by (I)

$\therefore$ $\alpha <\beta \Rightarrow$ (a)