Question

tan⁻¹ 1/2 + tan⁻¹ 1/5 + tan⁻¹ 1/8=π/4,Prove it.

Answer 1

LHS = tan^-1(1/2) + tan^-1(1/5) + tan^-1(1/8)

= [tan^-1(1/2) + tan^-1(1/8) ] + tan^-1(1/5)


we know, tan^-1x + tan^-1y = tan^-1{(x + y)/(1 - xy)}

so, tan^-1(1/2) + tan^-1(1/8) = tan^-1{1/2 + 1/8}/{1 - /1/2 × 1/8}

=tan^-1 {(4 + 1)/8}/{(16 - 1)/16}

= tan^-1{10/15}

= tan^-1{2/3}

now, tan^-1{1/2} + tan^-1{1/8} = tan^-1{2/3}

so, [tan^-1(1/2) + tan^-1(1/8) ] + tan^-1(1/5) = tan^-1{2/3} + tan^-1{1/5}

= tan^-1{(2/3 + 1/5)}/{1 - 2/3 × 1/5}

= tan^-1{(10 + 3)/15}/{(15 -2)/15}

= tan^-1{13/13}

= tan^-1{1} = π/4 = RHS

Answer 2

Hello,

Solution:

we know that

${tan}^{ - 1} x + {tan}^{ - 1} y = {tan}^{ - 1} ( \frac{x + y}{1 - xy} ) \\ \\ {tan}^{ - 1} \frac{1}{2} + {tan}^{ - 1} \frac{1}{5} = {tan}^{ - 1} ( \frac{ \frac{1}{2} + \frac{1}{5} }{1 - \frac{1}{10} } ) \\ \\ = {tan}^{ - 1} ( \frac{7}{9} ) \\ \\ {tan}^{ - 1} \frac{7}{9} + {tan}^{ - 1} \frac{1}{8} = {tan}^{ - 1} ( \frac{ \frac{7}{9} + \frac{1}{8} }{1 - \frac{7}{72} } ) \\ \\ = {tan}^{ - 1} ( \frac{ \frac{56 + 9}{72} }{ \frac{72 - 7}{72} } ) \\ \\ = {tan}^{ - 1} ( \frac{65 \times 72}{65 \times 72} ) \\ \\ = {tan}^{ - 1} (1) \\ \\ = \frac{\pi}{4} \\ \\ since \: \: \tan( \frac{\pi}{4} ) = 1$
hence proved that tan⁻¹ 1/2 + tan⁻¹ 1/5 + tan⁻¹ 1/8=π/4

Hope it helps you