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Question 160 of 175

Q.If A+B+C=180∘A+B+C=180^\circ, prove that tan⁡A2tan⁡B2+tan⁡B2tan⁡C2+tan⁡C2tan⁡A2=1\tan\dfrac{A}{2}\tan\dfrac{B}{2} + \tan\dfrac{B}{2}\tan\dfrac{C}{2} + \tan\dfrac{C}{2}\tan\dfrac{A}{2} = 1. OR Prove that lim⁡θ→0sin⁡θθ=1\displaystyle\lim_{\theta\to 0}\dfrac{\sin\theta}{\theta} = 1.

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2023Subjective· 5mImportance★★★★★
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Writing A+B2=90∘−C2\frac{A+B}2=90^\circ-\frac C2 turns tan⁡ ⁣(A+B2)\tan\!\left(\frac{A+B}2\right) into cot⁡C2\cot\frac C2; expanding the addition formula and clearing denominators produces exactly the required symmetric identity.

Since A+B+C=180∘A+B+C=180^\circ, we have A+B=180∘−CA+B=180^\circ-C, so:

A+B2=90∘−C2  ⟹  tan⁡(A+B2)=tan⁡(90∘−C2)=cot⁡C2\frac{A+B}{2} = 90^\circ-\frac C2 \implies \tan\left(\frac{A+B}{2}\right) = \tan\left(90^\circ-\frac C2\right) = \cot\frac C2

By the tangent addition formula:

tan⁡(A+B2)=tan⁡A2+tan⁡B21−tan⁡A2tan⁡B2\tan\left(\frac{A+B}{2}\right) = \frac{\tan\frac A2+\tan\frac B2}{1-\tan\frac A2\tan\frac B2}

Setting these equal:

tan⁡A2+tan⁡B21−tan⁡A2tan⁡B2=cot⁡C2=1tan⁡C2\frac{\tan\frac A2+\tan\frac B2}{1-\tan\frac A2\tan\frac B2} = \cot\frac C2 = \frac{1}{\tan\frac C2}

Cross-multiplying:

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