# Given that; `sin(pi/12) = (sqrt6-sqrt2)/4` and `cos(pi/12) = (sqrt6+sqrt2)/4` Show that `tanx = (1-cos2x)/(sin2x)` and deduce that to find `tan(pi/24)` .

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We have to show that `tanx = (1-cos2x)/(sin2x)`

We know the following in trigonometry.

`cos2x = 1-2sin^2x`

`sin2x = 2sinxcosx`

`RHS`

`= (1-cos2x)/(sin2x)`

`= (1-(1-2sin^2x))/(2sinxcosx)`

`= (2sin^2x)/(2sinxcosx)`

`= (sinx)/(cosx)`

`= tanx`

`= LHS`

*So the identity `tanx = (1-cos2x)/(sin2x)` is proved.*

It is given that;

`sin(pi/12) = (sqrt6-sqrt2)/4`

`cos(pi/12) = (sqrt6+sqrt2)/4`

Let `x = pi/24`

`tanx = (1-cos2x)/(sin2x)`

`tan(pi/24) = (1-cos((2pi)/24))/(sin((2pi)/24))`

`tan(pi/24) = (1-((sqrt6+sqrt2)/4))/((sqrt6-sqrt2)/4)`

`tan(pi/24) = (4-sqrt2-sqrt6)/(sqrt6-sqrt2)`

`tan(pi/24) = ((4-sqrt2-sqrt6)(sqrt6+sqrt2))/(6-2)`

`tan(pi/24) = (4sqrt6+4sqrt2-sqrt12-2-6-sqrt12)/4`

`tan(pi/24) = (4sqrt6+4sqrt2-2sqrt3-2-6-2sqrt3)/4`

`tan(pi/24) = (4sqrt6+4sqrt2-4sqrt3-8)/4`

`tan(pi/24) = (sqrt6-sqrt3+sqrt2-2)`

** So the answer is** `tan(pi/24) = (sqrt6-sqrt3+sqrt2-2)`

**Sources:**