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) ${tan}^{-1}\frac{2mn}{m^2-n^2}+{tan}^{-1}\frac{2pq}{p^2-q^2}={tan}^{-1}\frac{2MN}{M^2-N^2}$
where $M=mp-nq,N=np+mq$
(b) $\frac{2}{3}{tan}^{-1}\frac{3m{n}^2-m^3}{n^3-3m^{2}n}+\frac{2}{3}{tan}^{-1}\frac{3p{q}^2-p^3}{q^3-3p^{2}q}={tan}^{-1}\frac{2MN}{M^2-N^2}$
Where $M=-mp+nq,N=np+mq$ and all the letters are $+ive$ quantities.

Answer 1

(a) Dividing $T_1,T_2$ and $T_3$ by $m^2,p^2$ and $M^2$respectively, $2{tan}^{-1}x={tan}^{-1}\frac{2x}{1-x^2}$

L.H.S.$=2{tan}^{-1}\frac{n}{m}+2{tan}^{-1}\frac{q}{p}$

$=2{tan}^{-1}\frac{pn+mq}{pm-nq}=2{tan}^{-1}\frac{N}{M}=$R.H.S.

(b) Dividing above and below by $n^3$ in $T_1$ and $q_3$ in $T_2$ and $M^2$ in $T_3$, we get

$\frac{2}{3},3{tan}^{-1}\frac{m}{n}+\frac{2}{3},3{tan}^{-1}\frac{p}{q}=2{tan}^{-1}\frac{N}{M}$

L.H.S.$=2\left[{tan}^{-1}\frac{\left(m/n\right)+\left(p/q\right)}{1-\left(m/n\right)\left(p/q\right)}\right]$

$=2{tan}^{-1}\frac{np+mq}{nq-mp}=2{tan}^{-1}\frac{N}{M}=$R.H.S.