Prove that $\frac{(\sin 20^\circ + \cos 20^\circ)^2}{\cos 40^\circ} = \cot 25^\circ$

Notice that $\cos 40 =\sin 50$ and $ \sin 40 = \sin 50$. So we have \begin{eqnarray*} \frac{1+\sin 40}{ \cos 40} = \frac{1+\cos(50)}{\sin 50} \end{eqnarray*} Now use the double angle formulea $ \sin 2 \alpha = 2 \sin \alpha \cos \alpha$ and $ cos 2 \alpha =2 \cos^2 \alpha -1$. So \begin{eqnarray*} \frac{1+\cos(50)}{\sin 50} = \frac{1+2 \cos^2 25 -1}{2 \sin 25 \cos 25 } = \frac{\cos 25}{\sin 25} = \color{red}{ \cot 25}. \end{eqnarray*}

Expanding the square, using relationship $2 \sin a \cos a= \sin 2a$ and the definition of $\cot$, we have to show that

$$\frac{[(\sin20^\circ)^2 + (\cos20^\circ)^2] + \sin 40^\circ}{\cos40^\circ} = \dfrac{\cos25^\circ}{\sin25^\circ}$$

As the content of the square brackets is 1, it is equivalent to show that :

$$\sin25^\circ + \sin25^\circ \sin 40^\circ=\cos40^\circ\cos25^\circ$$ or

$$\sin25^\circ =\cos40^\circ\cos25^\circ - \sin25^\circ \sin 40^\circ$$

or $$\sin25^\circ =\cos(40^\circ+25^\circ)$$

or $$\sin25^\circ =\cos(90^\circ-65^\circ)$$

which is evidently true.