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Question Number 54239 by 951172235v last updated on 01/Feb/19

Answered by Prithwish sen last updated on 01/Feb/19

$$={\tan }^{-1}\left(\frac{\frac{1}{\mathrm{p}+\mathrm{q}+\mathrm{s}}+\frac{1}{\mathrm{p}+\mathrm{q}+\mathrm{t}}}{1-\frac{1}{(\mathrm{p}+\mathrm{q}+\mathrm{s})(\mathrm{p}+\mathrm{q}+\mathrm{t})}}\right)+$$$${\tan }^{-1}\left(\frac{\frac{1}{\mathrm{p}+\mathrm{r}+\mathrm{u}}+\frac{1}{\mathrm{p}+\mathrm{r}+\mathrm{v}}}{1-\frac{1}{(\mathrm{p}+\mathrm{r}+\mathrm{u})(\mathrm{p}+\mathrm{r}+\mathrm{v})}}\right)$$$$={\tan }^{-1}\left(\frac{2\mathrm{p}+2\mathrm{q}+\mathrm{s}+\mathrm{t}}{{(\mathrm{p}+\mathrm{q})}^2+(\mathrm{p}+\mathrm{q})(\mathrm{s}+\mathrm{t})+\mathrm{st}-1}\right)+$$$${\tan }^{-1}\left(\frac{2\mathrm{p}+2\mathrm{r}+\mathrm{u}+\mathrm{v}}{{(\mathrm{p}+\mathrm{r})}^2+(\mathrm{p}+\mathrm{r})(\mathrm{u}+\mathrm{v})+\mathrm{uv}-1}\right)$$$$={\tan }^{-1}\left(\frac{2\mathrm{p}+2\mathrm{q}+\mathrm{s}+\mathrm{t}}{2{(\mathrm{p}+\mathrm{q})}^2+(\mathrm{p}+\mathrm{q})(\mathrm{s}+\mathrm{t})}\right)+$$$${\tan }^{-1}\left(\frac{2\mathrm{p}+2\mathrm{r}+\mathrm{u}+\mathrm{v}}{2{(\mathrm{p}+\mathrm{r})}^2+(\mathrm{p}+\mathrm{r})(\mathrm{u}+\mathrm{v})}\right)$$$$={\tan }^{-1}\left(\frac{1}{(\mathrm{p}+\mathrm{q})}\right)+{\tan }^{-1}\left(\frac{1}{(\mathrm{p}+\mathrm{r})}\right)$$$$={\tan }^{-1}\left(\frac{2\mathrm{p}+\mathrm{q}+\mathrm{r}}{{\mathrm{p}}^2+\mathrm{pr}+\mathrm{pq}+\mathrm{qr}-1}\right)$$$$={\tan }^{-1}\left(\frac{2\mathrm{p}+\mathrm{q}+\mathrm{r}}{{2\mathrm{p}}^2+\mathrm{pq}+\mathrm{pr}}\right)$$$$={\tan }^{-1}\left(\frac{1}{\mathrm{p}}\right)$$$$\mathrm{Hence}\mathrm{proved}.$$

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