A simple equation involving trigonometry If $$\cos\theta+ \sin\theta + \tan\theta=\sqrt2+1$$
then what is the maximum value of$$\sin\theta,\sin^2\theta?$$

Let $a=\sin \theta,\ b=\cos \theta$. Then $a^2+b^2=1$ and 
$$a+b+\frac{a}{b}=\sqrt{2}+1$$
Notice that:
$$a+b\leq |a+b| \leq \sqrt{2(a^2+b^2)} = \sqrt{2}$$
and thus we deduce that $\dfrac{a}{b}\geq 1$. So $a,b$ have the same sign. Since we want the maximum of $a$, we are only interested in the case when $a$ and $b$ are positive. In this case, using AM-GM:
$$\sqrt{2}+1=a+b+\frac{1}{\sqrt{2}} \frac{a}{b}+\left(1-\frac{1}{\sqrt{2}}\right)\frac{a}{b}\geq 3\sqrt[3]{\frac{a^2}{\sqrt{2}}}+1-\frac{1}{\sqrt{2}}$$
and from here, it's easy to find that $a\leq \dfrac{1}{\sqrt{2}}$.
