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Miscellaneous Exercise 3 (II) · Q103

Q.Prove the following: 3tan⁡610°−27tan⁡410°+33tan⁡210°=13\tan^610°-27\tan^410°+33\tan^210°=1

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Step 1: Let t=tan⁡10°t=\tan10°. Since tan⁡30°=13=3t−t31−3t2\tan30°=\dfrac{1}{\sqrt3}=\dfrac{3t-t^3}{1-3t^2}: 3(3t−t3)=1−3t2\sqrt3(3t-t^3)=1-3t^2, i.e. 3t(3−t2)=1−3t2\sqrt3t(3-t^2)=1-3t^2.

Step 2: Rearrange to isolate the surd: 3t(3−t2)=1−3t2\sqrt3t(3-t^2)=1-3t^2; square both sides: 3t2(3−t2)2=(1−3t2)23t^2(3-t^2)^2=(1-3t^2)^2.

Step 3: Let u=t2u=t^2. Expand: 3u(9−6u+u2)=1−6u+9u23u(9-6u+u^2)=1-6u+9u^2, i.e. 27u−18u2+3u3=1−6u+9u227u-18u^2+3u^3=1-6u+9u^2. …

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