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**arze** By using the formulae expressing $\displaystyle sin\theta$ and $\displaystyle cos\theta$ in terms of t, where $\displaystyle (t\equiv tan\frac{\theta}{2})$ or otherwise, show that $\displaystyle \frac{1+sin\theta}{5+4cos\theta}\equiv\frac{(1+t)^ 2}{9+t^2}$.

Deduce that $\displaystyle 0\leq\frac{1+sin\theta}{5+4cos\theta}\leq\frac{10} {9}$

I have done the first part using the identities $\displaystyle sin\theta\equiv\frac{2t}{1+t^2}$ and $\displaystyle cos\theta\equiv\frac{1-t^2}{1+t^2}$ where $\displaystyle t\equiv tan\frac{\theta}{2}$, but I don't know how to do the second part.

Thanks