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) ${sin}^{-1}\frac{12}{13}+{cos}^{-1}\frac{4}{5}+{tan}^{-1}\frac{63}{16}=\pi$
(b) $2{cos}^{-1}\frac{3}{\sqrt{13}}+{cot}^{-1}\frac{16}{63}+\frac{1}{2}{cos}^{-1}\frac{7}{25}=\pi$

Answer 1

(a) Since ${sin}^{-1}\frac{12}{13}={tan}^{-1}\frac{12}{5}$
and ${cos}^{-1}\frac{4}{5}={tan}^{-1}\frac{3}{4}$, we have
L.H.S. $={tan}^{-1}\frac{12}{5}+{tan}^{-1}\frac{3}{4}+{tan}^{-1}\frac{63}{16}$
and since $\frac{12}{5}\times \frac{3}{4}>1$, we have
${tan}^{-1}\frac{12}{5}+{tan}^{-1}\frac{3}{4}=\pi +{tan}^{-1}\frac{\left(12/5\right)+\left(3/4\right)}{1-\left(12/5\right).\left(3/4\right)}$
$=\pi +{tan}^{-1}\{-\frac{63}{16}\}$
$=\pi +{tan}^{-1}\frac{63}{16}$ $[\because {tan}^{-1}(-x)=-{tan}^{-1}x]$
Hence
L.H.S. $=\pi -{tan}^{-1}\frac{63}{16}+{tan}^{-1}\frac{63}{16}=\pi =$ R.H.S.
(b) As in part (a), we will change each term to ${tan}^{-1}$
$2{cos}^{-1}\frac{3}{\sqrt{13}}={cos}^{-1}(2x^2-1)$
$={cos}^{-1}\frac{5}{13}={tan}^{-1}\frac{12}{13}$
${cot}^{-1}\frac{16}{63}={tan}^{-1}\frac{63}{16}$
$\frac{1}{2}{cos}^{-1}\frac{7}{25}=\theta$say, $\therefore$ $2\theta =\frac{7}{25}$
or $\frac{1-t^2}{1+t^2}=\frac{7}{25}$ $\therefore t^2=\frac{9}{16}$ or $t=\frac{3}{4},-\frac{3}{4}$
Since $cos2\theta =+ive$ $0\le 2\theta <\pi$
$\therefore$ $0\le \theta <\pi /2$
and hence $t=tan\theta =+ive$ $\therefore tan\theta =3/4$
$\therefore$ L.H.S. $={tan}^{-1}\frac{12}{13}+{tan}^{-1}\frac{63}{16}+{tan}^{-1}\frac{3}{4}$
Rest as in part (a).