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How to compute the following trigonometric question $$\sqrt2\sin10 (\sec5+\frac{2\cos 40}{\sin5}-4\sin35)=...$$ I am having problem to solve this trigonometric question. I tried to use identity $\sin10=2\sin5\cos5,\cos40=\cos(45-5), and \sin35=\sin(30+5)$ but it became complicated and I cannot simplify into a simpler term. I checked on Wolfram Alpha and it seemed the answer is 4 but I cannot get it. Could anyone here help me? Any help would be appreciated. Thanks in advance.

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(All the arguments are in degrees)

Let the expression to be evaluated be $x$.

Since $\sin 10=2\sin 5\cos 5$, you get: $$2\sqrt{2}(\sin 5+2\cos 40 \cos 5-4\sin 35\sin 5\cos 5)=2\sqrt{2}(\sin 5+2\cos 40 \cos 5-2\sin 35\sin 10)$$

We can write: $$2\cos 40 \cos 5=\cos 45+\cos 35$$ $$2\sin 35\sin 10=\cos 25-\cos 45$$ Hence, $$x=2\sqrt{2}(\sin 5+\cos 45+\cos 35+\cos 45-\cos 25)=2\sqrt{2}(\sqrt{2}+\sin 5 +\cos 35-\cos 25)$$ Since $$\cos 35-\cos 25=-2\sin 30\sin 5=-\sin 5$$ The final answer is $\boxed{4}$

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