No, my question is correct:
Show that the exact value of tan31°43' is \frac{\sqrt{5}-1}{2}
I got the solution to it.
Tan63°26' = 2
tanα = \frac{2tanα}{1-tan^{2}α}= 2
where (where α=31°43' (acute angle))
2t = 2-2t^{2}
t^{2}+t-1=0 , t>0
t= \frac{-1+\sqrt{5}-1}{2\times1} =...
1. Homework Statement
Show that the exact value of tan°43' is \frac{\sqrt{5}-1}{2}
2. Homework Equations
3. The Attempt at a Solution
tan2x = \frac{2tanx}{1-tan^{2}x}
\frac{2tan31° 43}{1-tan^{2}31° 43'}
From here, i got stuck or I'm doing it wrongly, i forgot how to do...