Question

tan60°.tan40°.tan20°=1/8


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Answer 1

$\huge \underline \mathfrak {Solution:-}$

LHS

$= \tan(60) \tan(40) \tan(20) \\ \\ = \sqrt{3} \times \frac{ \sin(40) }{ \cos(40) } \times \frac{ \sin(20) }{ \cos(20) } \\ \\ = \sqrt{3} \times \frac{2 \sin(40) \sin(20) }{2 \cos(40) \cos(20) } \\ \\ = \sqrt{3} \times \frac{ \cos(40 - 20) - \cos(40 + 20) }{ \cos(40 + 20) + \cos(4 0 - 20) } \\ \\ = \sqrt{3} \times \frac{ \cos(20) - \cos(60) }{ \cos(60) + \cos(20) } \\ \\ = \sqrt{3} \times \frac{ \cos(20) - \frac{1}{2} }{ \frac{1}{2} + \cos(20) } \\ \\ = \sqrt{3} \times \frac{2 \cos(20) - 1 }{2 \cos(20) + 1}$

Hope that there must be 80 in place of 40

Answer 2

Step-by-step explanation:

★ Correction is required in the Question .

Q. Prove that tan20° tan40° tan60° tan80°= 3

Sol :

➤ tan20°tan40°tan60°tan80°

➤ {(sin20°sin40°sin80°)/(cos20°cos40°cos80°)}×√3

➤ √3×{(2sin20°sin40°sin80°)/(2cos20°cos40°cos80°)}

➤ √3×[{cos(20°-40°)-cos(20°+40°)}sin80°/{cos(20°-40°)+cos(20°+40°)}cos80°]

➤ √3×[{cos20°sin80°-(1/2)sin80°}/{cos20°cos80°+(1/2)cos80°}] [∵, cos60°=1/2]

➤ √3×[{sin100°-sin(-60°)-sin80°}/{cos60°+cos100°+cos80°}]

➤ √3×[{sin100°-sin80°+√3/2}/{1/2+cos100°+cos80°}]

➤ √3×[{2cos(100°+80°)/2sin(100°-80°)/2+√3/2}/{1/2+2cos(100°+80°)/2cos(100°-80°)/2}]

➤ √3×√3/2×2/1

➤ 3 . Hence Proved