Question

Question #adbd7

Answer 1

`tan3a=tan(2a+a)`

`=>tan3a=(tan2a+tana)/(1-tan2atana`

`=>tan3a=((2tana)/(1-tan^2a)+tana)/(1-(2tana)/(1-tan^2a)tana`

`=>tan3a=(2tana+(1-tan^2a)tana)/(1-tan^2a-2tan^2a)`

`=>tan3a=(3tana-tan^3a)/(1-3tan^2a)`

`=>tan3a=(tana(3-tan^2a))/(1-3tan^2a)`

`=(tana(sqrt3+tana)(sqrt3-tana))/((1-sqrt3tana)(1+sqrt3tana))`

`=(tana(tan60^@+tana)(tan60^@-tana))/((1-tan60^@tana)(1+tan60^@tana))`

`=>color(red)(tan3a=tanatan(60^@+a)tan(60^@-a))...[1]`

Inserting `a=6^@` in [1] we get

`color(blue)(tan18^@=tan(6^@)tan(66^@)tan(54^@)....[2])`

Inserting `a=18^@` in [1] we get

`color(green)(tan54^@=tan(18^@)tan(78^@)tan(42^@)....[3])`

Multiplying [2] by [3] we get

`color(magenta)(tan(6^@)tan(42^@)tan(66^@)tan(78^@)=1)`