Question

if tan(alfa+beta)=root3 and tan(alfa--beta)=1then6bete=?​

Answer 1

Answer:

$\tan( \alpha + \beta ) = \sqrt{3} \\ \implies \: \tan( \alpha + \beta ) = tan \: {60}^{0} \\ \implies \: \alpha + \beta = {60}^{0} .....(1) \\ \\ \tan( \alpha - \beta ) = 1 \\ \implies \: \tan( \alpha - \beta ) = tan \: {45}^{0} \\ \implies \: \alpha - \beta = {45}^{0} .....(2) \\ substracting \: (2) \: from \: (1) \: \: \\ we \: find \\ \alpha + \beta - ( \alpha - \beta ) = {60}^{0} - {45}^{0} \\ 2 \beta \: = \: {15}^{0} \\ 3 \times 2 \beta \: = 3 \times \: {15}^{0} \\ 6 \beta \: = {45}^{0}$