Question

If $p>q>0$ and $pr<-1<qr$, then $ta{n}^{-1}\left(\frac{p-q}{1+pq}\right)+ta{n}^{-1}\left(\frac{q-r}{1+qr}\right)+ta{n}^{-1}\left(\frac{r-p}{1+rp}\right)=$_.

• A. $\pi$
• B. $2\pi$
• C. $\frac{\pi }{2}$
• D. $\frac{3\pi }{2}$

Answer 1

Answer: (A)

Since, $p$, $q>0$ therefore $pq>0$
and $ta{n}^{-1}\left(\frac{p-q}{1+pq}\right)=ta{n}^{-1}p-ta{n}^{-1}q\dots (i)$
Since, $qr>-1$
$\therefore ta{n}^{-1}\left(\frac{q-r}{1+qr}\right)=ta{n}^{-1}q-ta{n}^{-1}r\dots (ii)$
and since $pr<-1$ and $r<0$
$\therefore ta{n}^{-1}\left(\frac{r-p}{1+rp}\right)=\pi +ta{n}^{-1}r-ta{n}^{-1}p\dots (iii)$
On adding Eqs. $\left(i\right)$, $\left(ii\right)$ and $\left(iii\right)$, we get
$ta{n}^{-1}\left(\frac{p-q}{1+pq}\right)+ta{n}^{-1}\left(\frac{q-r}{1+qr}\right)+ta{n}^{-1}\left(\frac{r-p}{1+rp}\right)$
$=\pi$