Question

Question #90557

Answer 1

`tan(pi/8)=(sqrt2-1).`

Explanation:

We have,

`tan^2(theta/2)=sin^2(theta/2)/cos^2(theta/2)xx2/2,`

`={2sin^2(theta/2)}/{2cos^2(theta/2)},`

`=(1-costheta)/(1+costheta),`

`:. tan(theta/2)=pmsqrt{(1-costheta)/(1+costheta)}.`

Letting, `theta=pi/4," so that, "theta/2=pi/8,` we have,

`tan(pi/8)=pmsqrt{(1-cos(pi/4))/(1+cos(pi/4))},`

`=pmsqrt{(1-1/sqrt2)/(1+1/sqrt2)},`

`=pmsqrt[{(sqrt2-1)/(sqrt2+1)}xx{(sqrt2-1)/(sqrt2-1)}],`

`=pmsqrt[{(sqrt2-1)^2)/(2-1)],`

`:. tan(pi/8)=pm(sqrt2-1).`

As `(pi/8)" lies in the "1^(st)` Quadrant, it is `+ve.`

`rArr tan(pi/8)=+(sqrt2-1).`

Enjoy Maths.!