Originally Posted by

**Grandad** Hello thereddevilsThanks for showing us your working. But I think there must be something wrong with the question. For any angle $\displaystyle A < 30^o$, the cosines of $\displaystyle A, 2A$ and $\displaystyle 3A$ are all positive. So the result can't possibly equal $\displaystyle -\cot^2(\tfrac12A)$, can it?

When you've got as far as $\displaystyle \frac{2\cos A+2}{2\cos A+1}$, did you think about using the half-angle formula using $\displaystyle t = \tan(\tfrac12A)$? Then $\displaystyle \cos A = \frac{1-t^2}{1+t^2}$

So $\displaystyle \frac{2\cos A+2}{2\cos A+1}= \frac{\dfrac{2(1-t^2)}{1+t^2}+2}{\dfrac{2(1-t^2)}{1+t^2}+1} = ...$ ?

Grandad