Question

show that 4tan inverse 1/5-tan inverse 1/70+tan inverse 1/99=π\4

Answer 1

Please see the attachment

Answer 2

Answer:

Since, $2tan^{-1}x = tan \frac{2x}{1-x^2}$

And,

$tan^{-1}x+tan^{-1}y=tan^{-1}\frac{x+y}{1-xy}$

We have to prove:

$4tan^{-1}\frac{1}{5}-tan^{-1}\frac{1}{70}+tan^{-1}\frac{1}{99}=\frac{\pi}{4}$

L.H.S.

$4tan^{-1}\frac{1}{5}-tan^{-1}\frac{1}{70}+tan^{-1}\frac{1}{99}$

$2(2tan^{-1}\frac{1}{5}) -[tan^{-1}\frac{1}{70}-tan^{-1}\frac{1}{99}]$

$2tan^{-1}(\frac{2\times 1/5}{1-1/25})-tan^{-1} [ \frac{1/70-1/99}{1+1/70\times 1/99}]$

$2tan^{-1}(\frac{2/5}{24/25})-tan^{-1} [ \frac{29/6930}{6931/6930}]$

$2tan^{-1}\frac{5}{12}-tan^{-1} (\frac{29}{6931})$

$tan^{-1}(\frac{2\times 5/12}{1-25/144})-tan^{-1}\frac{1}{239}$

$tan^{-1}\frac{10/12}{119/144}-tan^{-1}\frac{1}{239}$

$tan^{-1}\frac{120}{119}-tan^{-1}\frac{1}{239}$

$tan^{-1}[\frac{120/119-1/239}{1+120/119\times 1/239}]$

$tan^{-1}[\frac{28680-119}{28441+120}]$

$tan^{-1}\frac{28561}{28561}]=tan^{-1}(1) = \frac{\pi}{4}$

= R.H.S.

Hence, proved.