Question-54239

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

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

= \tan^{- 1} (\frac{\frac{1}{p + q + s} + \frac{1}{p + q + t}}{1 - \frac{1}{(p + q + s) (p + q + t)}}) +

\tan^{- 1} (\frac{\frac{1}{p + r + u} + \frac{1}{p + r + v}}{1 - \frac{1}{(p + r + u) (p + r + v)}})

= \tan^{- 1} (\frac{2 p + 2 q + s + t}{{(p + q)}^{2} + (p + q) (s + t) + st - 1}) +

\tan^{- 1} (\frac{2 p + 2 r + u + v}{{(p + r)}^{2} + (p + r) (u + v) + uv - 1})

= \tan^{- 1} (\frac{2 p + 2 q + s + t}{2 {(p + q)}^{2} + (p + q) (s + t)}) +

\tan^{- 1} (\frac{2 p + 2 r + u + v}{2 {(p + r)}^{2} + (p + r) (u + v)})

= \tan^{- 1} (\frac{1}{(p + q)}) + \tan^{- 1} (\frac{1}{(p + r)})

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

= \tan^{- 1} (\frac{2 p + q + r}{{2 p}^{2} + pq + pr})

= \tan^{- 1} (\frac{1}{p})

Hence proved .