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## Homework Statement

Show that the exact value of tan°43' is [itex]\frac{\sqrt{5}-1}{2}[/itex]

## Homework Equations

## The Attempt at a Solution

[itex]tan2x = \frac{2tanx}{1-tan^{2}x}[/itex]

[itex]\frac{2tan31° 43}{1-tan^{2}31° 43'}[/itex]

From here, i got stuck or I'm doing it wrongly, i forgot how to do it.

Btw, I'm new here, please tell me if i posted it in the wrong place xD