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Exercise 3.1 · Q3

Q.Find the value of tan⁡105°\tan 105°

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Step 1: Write 105°=60°+45°105°=60°+45° and use tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}.

Step 2: tan⁡105°=tan⁡60°+tan⁡45°1−tan⁡60°tan⁡45°=3+11−3\tan105°=\dfrac{\tan60°+\tan45°}{1-\tan60°\tan45°}=\dfrac{\sqrt3+1}{1-\sqrt3}.

Step 3: Rationalise by multiplying numerator and denominator by (1+3)(1+\sqrt3): (3+1)(1+3)(1−3)(1+3)=(3+1)21−3=4+23−2\dfrac{(\sqrt3+1)(1+\sqrt3)}{(1-\sqrt3)(1+\sqrt3)}=\dfrac{(\sqrt3+1)^2}{1-3}=\dfrac{4+2\sqrt3}{-2}.

Step 4: Simplify: 4+23−2=−(2+3)\dfrac{4+2\sqrt3}{-2}=-(2+\sqrt3).

✓Final answer

tan⁡105°=−(2+3)\tan105°=-(2+\sqrt3)

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