Question

Prove that :-



${ \boxed{ \bf{ \tan { }^{ - 1} \frac{1}{5} + \tan {}^{ - 1} \frac{1}{7} + \tan {}^{ - 1} \frac{1}{3} + \tan {}^{ - 1} \frac{1}{8} = \frac{ \pi}{4} }}}$
​

Answer 1

Answer:

Your answer attached in the photo

Answer 2

TO PROVE :–

$\\ \implies\bf{ \tan^{ - 1} \dfrac{1}{5} + \tan^{ - 1} \dfrac{1}{7} + \tan^{ - 1} \dfrac{1}{3} + \tan^{ - 1} \dfrac{1}{8} = \dfrac{\pi}{4}}$

SOLUTION :–

• Let's take L.H.S. –

$\\ \: \: = \: \: \bf\tan^{ - 1} \dfrac{1}{5} + \tan^{ - 1} \dfrac{1}{7} + \tan^{ - 1} \dfrac{1}{3} + \tan^{ - 1} \dfrac{1}{8}$

• We know that –

$\\\implies \large \red{ \boxed{\bf\tan^{ - 1}(x) + \tan^{ - 1}(y) = { \tan}^{ - 1} \left[ \dfrac{x + y}{1 - xy} \right]}}$

• So that –

$\\ \: \: = \: \: \bf{ \tan}^{ - 1} \left[ \dfrac{\dfrac{1}{5} + \dfrac{1}{7}}{1 -\dfrac{1}{5} \times \dfrac{1}{7}}\right] +{ \tan}^{ - 1} \left[ \dfrac{\dfrac{1}{3} + \dfrac{1}{8}}{1 -\dfrac{1}{3} \times \dfrac{1}{8}}\right]$

$\\ \: \: = \: \: \bf{ \tan}^{ - 1} \left[ \dfrac{\dfrac{7 + 5}{5 \times 7}}{\dfrac{35 - 1}{5 \times 7}}\right] +{ \tan}^{ - 1} \left[ \dfrac{\dfrac{8 + 3}{3 \times 8}}{\dfrac{24 - 1}{3 \times 8}}\right]$

$\\ \: \: = \: \: \bf{ \tan}^{ - 1} \left[ \dfrac{12}{34}\right] +{ \tan}^{ - 1} \left[ \dfrac{11}{23}\right]$

• Again using formula –

$\\ \: \: = \: \: \bf{ \tan}^{ - 1} \left[ \dfrac{ \dfrac{12}{34} + \dfrac{11}{23}}{1 - \dfrac{12}{34} \times \dfrac{11}{23}} \right]$

$\\ \: \: = \: \: \bf{ \tan}^{ - 1} \left[ \dfrac{ \dfrac{12 \times 23 + 11 \times 34}{34 \times 23}}{ \dfrac{34 \times 23 - 12 \times 11}{34 \times 23} }\right]$

$\\ \: \: = \: \: \bf{ \tan}^{ - 1} \left[ \dfrac{ 12 \times 23 + 11 \times 34}{34 \times 23 - 12 \times 11}\right]$

$\\ \: \: = \: \: \bf{ \tan}^{ - 1} \left[ \dfrac{276+374}{782- 132}\right]$

$\\ \: \: = \: \: \bf{ \tan}^{ - 1} \left[ \dfrac{650}{650}\right]$

$\\ \: \: = \: \: \bf{ \tan}^{ - 1}(1)$

$\\ \bf\: \: = \: \: \dfrac{\pi}{4}$

$\\ \bf\: \: = \: \:R.H.S.$

$\\\longrightarrow \large \green{\boxed{ \bf Hence \: \: Proved}}$