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) $sin\left(2{tan}^{-1}\frac{1}{3}\right)+cos\left({tan}^{-1}2\sqrt{2}\right)=\frac{14}{15}$
(b) Prove that
${sin}^{-1}\left\{cot({sin}^{-1}\sqrt{\left(\frac{2-\sqrt{3}}{4}\right)}+{cos}^{-1}\frac{\sqrt{12}}{4}+{sec}^{-1}\sqrt{2})\right\}=0$

Answer 1

(a) $2{tan}^{-1}x={sin}^{-1}\frac{2x}{1+x^2}$

${tan}^{-1}y={cos}^{-1}\frac{1}{\surd (1+y^2)}$

$L.H.S.=\frac{2x}{1+x^2}+\frac{1}{\sqrt{(1+y^2)}},x=\frac{1}{3},y=2\sqrt{2}$

$\frac{3}{5}+\frac{1}{3}=\frac{14}{15}$

(b) ${sec}^{-1}\sqrt{2}={cos}^{-1}\frac{1}{\sqrt{2}}=\frac{\pi }{6}={45}^0$

${cos}^{-1}\frac{\sqrt{12}}{4}={cos}^{-1}\frac{\sqrt{3}}{2}=\frac{\pi }{6}={30}^0$

${sin}^{-1}\sqrt{\frac{2-\sqrt{3}}{4}}={sin}^{-1}\sqrt{\frac{4-2\sqrt{3}}{8}}$

$={sin}^{-1}\frac{\sqrt{3}-1}{2\sqrt{2}}={15}^0$

$\therefore$ $L.H.S.={sin}^{-1}\left[cot\right({45}^0+{30}^0+{15}^0\left)\right]$

$={sin}^{-1}\left(cot{90}^0\right)={sin}^{-1}0=0$

$\therefore$ $L.H.S.=0$.