Question

Prove that 2 tan⁻¹ 1/2 + tan⁻¹ 1/7 = tan⁻¹ 31/17.

Answer 1

Formula used:

$2 \: {tan}^{ - 1} x = {tan}^{ - 1} ( \frac{2x}{1 - {x}^{2} } ) \\ \\ {tan}^{ - 1} x + {tan}^{ - 1} y = {tan}^{ - 1} ( \frac{x + y)}{1 - xy)}$
So,

$2 \: {tan}^{ - 1} ( \frac{1}{2} ) = {tan}^{ - 1} ( \frac{2 \times \frac{1}{2} }{1 - {( \frac{1}{2} )}^{2} } ) \\ \\ = {tan}^{ - 1} ( \frac{1}{1 - \frac{1}{4} } ) \\ \\ 2 {tan}^{ - 1} ( \frac{1}{2} )= {tan}^{ - 1} ( \frac{4}{3} ).....eq1$

${tan}^{ - 1} ( \frac{4}{3} ) + {tan}^{ - 1} ( \frac{1}{7} ) = {tan}^{ - 1} ( \frac{ \frac{4}{3} + \frac{1}{7} }{1 - \frac{4 }{21} } ) \\ \\ \\ = {tan}^{ - 1} ( \frac{ \frac{28 + 3}{21} }{ \frac{21 - 4}{21} } ) \\ \\ \\ = {tan}^{ - 1} ( \frac{31}{17} )$
= RHS

HENCE PROVED