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Question 34 of 36

Q.If A=30°A=30° then prove that sin⁡2A=2tan⁡A1+tan⁡2A\sin 2A=\dfrac{2\tan A}{1+\tan^2 A}

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Commerce Board 2025Subjective· 2mImportance★★★★★
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With A=30∘A=30^\circ: LHS =sin⁡60∘=32=\sin 60^\circ=\tfrac{\sqrt3}{2}; RHS =2tan⁡30∘1+tan⁡230∘=32=\dfrac{2\tan 30^\circ}{1+\tan^2 30^\circ}=\tfrac{\sqrt3}{2}. LHS == RHS.

To prove (for A=30∘A=30^\circ): sin⁡2A=2tan⁡A1+tan⁡2A\sin 2A=\dfrac{2\tan A}{1+\tan^2 A}.

Step 1 — evaluate the LHS.

LHS=sin⁡2A=sin⁡(2×30∘)=sin⁡60∘=32.\text{LHS}=\sin 2A=\sin(2\times30^\circ)=\sin 60^\circ=\frac{\sqrt3}{2}.

Step 2 — evaluate the RHS using tan⁡30∘=13\tan 30^\circ=\dfrac{1}{\sqrt3}.

RHS=2tan⁡30∘1+tan⁡230∘=2⋅131+13=2343=23×34=643=323.\text{RHS}=\frac{2\tan 30^\circ}{1+\tan^2 30^\circ}=\frac{2\cdot\dfrac{1}{\sqrt3}}{1+\dfrac{1}{3}}=\frac{\dfrac{2}{\sqrt3}}{\dfrac{4}{3}}=\frac{2}{\sqrt3}\times\frac{3}{4}=\frac{6}{4\sqrt3}=\frac{3}{2\sqrt3}.

Rationalise: …

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