Dear student cos10° cos30° cos50° cos70°=32×cos10° cos50°cos50°=34×2cos10° cos50°cos70°=34cos50°+10°+cos(50°-10°)cos70° ∵2cosA cosB=cos(A+B)+cos(A-B)=34cos60°+cos40°cos70°=3412+cos40°cos70°=3412cos70°+cos70° cos40°=38cos70°+2cos70° cos40°=38cos70°+cos(70°+40°+cos(70°-40°)) ∵2cosA cosB=cos(A+B)+cos(A-B)=38cos70°+cos110°+cos30°=38cos70°+cos110°+32=38cos70°+cos(180°-70°)+32 ∵cos(180°-x)=-cosx=38cos70°-cos70°+32=38×32=316 For remaining queries we request you to post them in separate threads to have rapid assistance from our experts Regards

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\cos 10 ° \cos 30 ° \cos 50 ° \cos 70 ° = \frac{\sqrt{3}}{2} \times (\cos 10 ° \cos 50 °) \cos 50 ° = \frac{\sqrt{3}}{4} \times (2 \cos 10 ° \cos 50 °) \cos 70 ° = \frac{\sqrt{3}}{4} [\cos (50 ° + 10 °) + \cos (50 ° - 10 °)] \cos 70 ° [∵ 2 cosA cosB = \cos (A + B) + \cos (A - B)] = \frac{\sqrt{3}}{4} (\cos 60 ° + \cos 40 °) \cos 70 ° = \frac{\sqrt{3}}{4} (\frac{1}{2} + \cos 40 °) \cos 70 ° = \frac{\sqrt{3}}{4} (\frac{1}{2} \cos 70 ° + \cos 70 ° \cos 40 °) = \frac{\sqrt{3}}{8} (\cos 70 ° + 2 \cos 70 ° \cos 40 °) = \frac{\sqrt{3}}{8} (\cos 70 ° + \cos (70 ° + 40 °) + \cos (70 ° - 40 °)) [∵ 2 cosA cosB = \cos (A + B) + \cos (A - B)] = \frac{\sqrt{3}}{8} (\cos 70 ° + \cos 110 ° + \cos 30 °) = \frac{\sqrt{3}}{8} (\cos 70 ° + \cos 110 ° + \frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{8} (\cos 70 ° + \cos (180 ° - 70 °) + \frac{\sqrt{3}}{2}) [∵ \cos (180 ° - x) = - cosx] = \frac{\sqrt{3}}{8} (\cos 70 ° - \cos 70 ° + \frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{8} \times \frac{\sqrt{3}}{2} = \frac{3}{16}

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