Proof tan 20° tan 35° tan 45° tan 55° tan 70° = 1
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L.H.S = tan 20° tan 35° tan 45° tan 55° tan 70° = tan (90° − 70°) tan (90° − 55°) tan 45°tan 55° tan 70° = cot 70°cot 55° tan 45° tan 55° tan 70° [∵ tan (90 – θ) = cot θ] = (tan 70° cot 70°)(tan 55° cot 55°) tan 45° [∵ tan θ x cot θ = 1] = 1 x 1 x 1 = 1 Hence p
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