Question

Question #8977f

Answer 1

`1 + sqrt2`

Explanation:

Call tan ( 67.5) = tan t.
tan (2t) = tan (135) = - 1
Use trig identity:
`tan 2t = (2tan t)/(1 - tan^2 t`
In this case:
`-1 = (2tan t)/(1 - tan^2 t)`
Cross multiply and bring quadratic equation to standard form:
`tan^2 t - 2tan t - 1 = 0`
`D = d^2 = b^2 - 4ac = 4 + 4 = 8` --> `d = +- 2sqrt2`
There are 2 real roots:
`tan t = -b/(2a) +- d/(2a) = 2/2 +- (2sqrt2)/2 = 1 +- sqrt2`
Since tan (67.5) > 0, there for:
`tan t = tan (67.5) = 1 + sqrt2`