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# Doubt from trigonometry

How sin80 have came in third step, which formulae is applied by you?

Identity is

sinA+sinB=2sin\frac{A+B}2cos\frac{A-B}2

and then

cos(90-\theta)=sin\theta

**Proof Without Words::**

Given \cot(70^\circ)+4\cos(70^\circ)=\tan(20^\circ)+4\sin(20^\circ)

From Diagram

\Longrightarrow \text{AC=AE+CE=AF+CE.}

\Longrightarrow\text{AC=2AD+CE}=4\sin(20^\circ)+\tan(20^\circ).

\Longrightarrow 4\sin(20^\circ)+\tan(20^\circ)=\text{AC}=\sqrt{3}.

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