Math, asked by Swapnonil, 1 year ago

tanAtan(60°+A)tan(60°-A)=tan3A

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Answered by Anonymous
150
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Answered by probrainsme104
10

Concept

Trigonometry could be a branch of mathematics that studies relationships between side lengths and angles of triangles.

Given

The given expression is\tan A\cdot \tan \left(60^{\circ}+A\right)\cdot \tan \left(60^{\circ}-A\right)=\tan3A.

Find

We have to prove the given expression.

Solution

We will prove the given expression by taking LHS.

We will substitute \tan \theta=\frac{\sin \theta}{\cos \theta}, we get

\frac{\sin A}{\cos A}\times \frac{\sin(60+A)}{\cos(60+A)}\times\frac{\sin (60-A)}{\cos(60-A)}

Further, we are going to use the identity \sin(A-B)\sin(A+B)=\sin^2 A-\sin^2 B and \cos(A-B)\cos(A+B)=\cos^2 A-\sin^2 B, we get

\frac{\sin A}{\cos A}\times \frac{\sin^2 60-\sin^2 A}{\cos^2 60-\sin^2 A}

As we know, \sin 60=\frac{\sqrt{3}}{2} and \cos 60=\frac{1}{2}.

\begin{aligned}\frac{\sin A}{\cos A}\times \frac{\frac{3}{4}-\sin^2 A}{\frac{1}{4}-\sin^2 A}\\ \frac{\sin A}{\cos A}\times \frac{3-4\sin^2 A}{1-4\sin^2 A}\end{aligned}

Now, we are going to use the identity \sin^2 \theta =1-\cos^2 \theta

\begin{aligned}&\frac{\sin A}{\cos A}\times \frac{3-4\sin^2 A}{1-4(1-\cos^2 A)}\\ &\frac{3\sin A-4\sin^3 A}{4\cos^2 A-3\cos A}\end{aligned}

Further, we are going to use the identity 3\sin A-4\sin^3 A=\sin3 A and 4\cos^2 A-3\cos A=\cos 3A

Therefore, \frac{\sin3A}{\cos 3A}=\tan 3A

Hence, the given expression is proved.

#SPJ2

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