<s>[INST] Prove that $\frac{(\sin 20^\circ + \cos 20^\circ)^2}{\cos 40^\circ} = \cot 25^\circ$ So I'm trying to come up with an answer to this question for hours now. I don't know what I'm doing wrong and none of the calculators on the internet couldn't help so I figured I should ask people.
What have I done so far:
$\frac{(\sin 20^\circ + \cos 20^\circ)^2}{\cos 40^\circ} = \frac{(\frac{2\sin(45^\circ+20^\circ)}{\sqrt{2}})^2}{\cos 40^\circ} = \frac{\sin^2 65^\circ}{\cos 40^\circ} = \frac{1 - \cos 130^\circ}{\cos 40^\circ} = \frac{1 + \cos 50^\circ}{\cos 40^\circ} = \frac{2\cos^2 25^\circ}{\cos 40^\circ}$ ... etc. 
I can't seem to figure out where to go from here so I'm just stuck.

I also tried the classic approach: 
$\frac{(\sin 20^\circ + \cos 20^\circ)^2}{\cos 40^\circ} = \frac{1 + \sin 40^\circ}{\cos 40^\circ} = \frac{1}{\cos 40^\circ} + {\tan 40^\circ} = \sec 40^\circ + \tan 40^\circ$ 
But how can I prove that
$\sec 40^\circ + \tan 40^\circ = \cot 25^\circ$ ?
What am I doing wrong? Any hints or solutions would be great. Thanks in advance.
 [/INST] 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
$$\sin25^\circ =\cos(65^\circ)$$
which is evidently true.
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