Question

Prove the following statement

${tan}^{ - 1} \frac{1}{1 + 1 + {1}^{2} } + {tan}^{ - 1} \frac{1}{1 + 2 + {2}^{2} } + ... + {tan}^{ - 1} \frac{1}{1 + n + {n}^{2}} = {tan}^{ - 1} (n + 1) - \frac{\pi}{4}$
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Answer 1

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm \: {tan}^{ - 1} \dfrac{1}{1 + 1 + {1}^{2} } + {tan}^{ - 1} \dfrac{1}{1 + 2 + {2}^{2} } + ... + {tan}^{ - 1} \dfrac{1}{1 + n + {n}^{2}}$

can be rewritten as

$\rm \: = {tan}^{ - 1} \dfrac{1}{1 + 2} + {tan}^{ - 1} \dfrac{1}{1 + 6} + ... + {tan}^{ - 1} \dfrac{1}{1 + n(n + 1)}$

$\rm \: = {tan}^{ - 1} \dfrac{1}{1 + 2.1} + {tan}^{ - 1} \dfrac{1}{1 + 3.2} + ... + {tan}^{ - 1} \dfrac{1}{1 + n(n + 1)}$

$\rm \: = {tan}^{ - 1} \dfrac{2 - 1}{1 + 2.1} + {tan}^{ - 1} \dfrac{3 - 2}{1 + 3.2} + ... + {tan}^{ - 1} \dfrac{n + 1 - n}{1 + n(n + 1)}$

We know,

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

So, using this result, we get

$\rm \: = ({tan}^{ - 1}2 - {tan}^{ - 1}1) + ({tan}^{ - 1}3 - {tan}^{ - 1}2) + \cdots \: + [{tan}^{ - 1}(n + 1) - {tan}^{ - 1}n]$

$\rm \: = \: {tan}^{ - 1}(n + 1) - {tan}^{ - 1}1$

$\rm \: = \: {tan}^{ - 1}(n + 1) - \dfrac{\pi}{4}$

Hence, Proved

$\rule{190pt}{2pt}$

Additional Information :-

$\boxed{ \rm{ \:{tan}^{ - 1}x + {tan}^{ - 1}y = {tan}^{ - 1} \frac{x + y}{1 - xy} \: provided \: that \: xy \leqslant 1 \: }}$

$\boxed{ \rm{ \:{tan}^{ - 1}x + {tan}^{ - 1}y = \pi + {tan}^{ - 1} \frac{x + y}{1 - xy} \: provided \: that \: xy > 1 1 \: }}$

$\boxed{ \rm{ \:2{tan}^{ - 1}x = {tan}^{ - 1} \frac{2x}{1 - {x}^{2} } \: }}$

$\boxed{ \rm{ \:2{tan}^{ - 1}x = {sin}^{ - 1} \frac{2x}{1 + {x}^{2} } \: }}$

$\boxed{ \rm{ \:2{tan}^{ - 1}x = {cos}^{ - 1} \frac{1 - {x}^{2} }{1 + {x}^{2} } \: }}$

$\boxed{ \rm{ \:{sin}^{ - 1}x + {sin}^{ - 1}y = {sin}^{ - 1}[x \sqrt{1 - {y}^{2} } + y \sqrt{1 - {x}^{2} } \: }}$

$\boxed{ \rm{ \:{sin}^{ - 1}x - {sin}^{ - 1}y = {sin}^{ - 1}[x \sqrt{1 - {y}^{2} } - y \sqrt{1-{x}^{2} }\:}}$