Question

If $p>q>0andpr<-1<qr$, then find the value of

${\tan }^{-1}\frac{p-q}{1+pq}+{\tan }^{-1}\frac{q-r}{1+qr}+{\tan }^{-1}\frac{r-p}{1+rp}.$

• A. $2\pi$
• B. $\pi$
• C. $3\pi$
• D. 0

Answer 1

Answer: (B)

Given:

$p>q>0$ and $pr<-1<qr$

$\Rightarrow {\tan }^{-1}p-{\tan }^{-1}q={\tan }^{-1}\frac{p-q}{1+pq}$ --(1)

${\tan }^{-1}q-{\tan }^{-1}r={\tan }^{-1}\frac{q-r}{1+qr}$ --(2)

${\tan }^{-1}p-{\tan }^{-1}r=-\pi +{\tan }^{-1}\frac{p-r}{1+rp}$ $\because pr<-1$

$\Rightarrow {\tan }^{-1}r-{\tan }^{-1}p=\pi +{\tan }^{-1}\frac{r-p}{1+rp}$ ---(3)

Now,

${\tan }^{-1}\frac{p-q}{1+pq}+{\tan }^{-1}\frac{q-r}{1+qr}+{\tan }^{-1}\frac{r-p}{1+rp}.$

$={\tan }^{-1}p-{\tan }^{-1}q+{\tan }^{-1}q-{\tan }^{-1}r+\pi +{\tan }^{-1}r-{\tan }^{-1}p$

$=\pi$

$\therefore {\tan }^{-1}\frac{p-q}{1+pq}+{\tan }^{-1}\frac{q-r}{1+qr}+{\tan }^{-1}\frac{r-p}{1+rp}=\pi$

Hence, Option C is correct.