Question

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

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