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Cos10° cos30° cos50° cos70° = 3/16

Answer» cos 10° cos 30° cos 50° cos 70° =\xa0{tex}\\frac{3}{16}{/tex}LHS = cos 10° cos 30° cos 50° cos 70°= cos 30° cos 10° cos 50° cos 70°{tex}=\\frac{\\sqrt{3}}{2}{/tex}\xa0(cos 10° cos 50° cos 70°){tex}=\\frac{\\sqrt{3}}{2}{/tex}\xa0(cos 10° cos 50°) cos 70°{tex}=\\frac{\\sqrt{3}}{4}{/tex}\xa0(2 cos 10° cos 50°) cos 70° [Multiplying and dividing by 2]Also,{tex}\\Rightarrow{/tex}\xa02 cos A cos B = cos (A + B) + cos (A - B) ...(i){tex}=\\frac{\\sqrt{3}}{4}{/tex}\xa0cos 70° {cos (50° + 10°) + cos (10° - 50°)}{tex}=\\frac{\\sqrt{3}}{4}{/tex}\xa0cos 70° {cos 60° + cos (-40°)}Now,Cos (-x) = cos x{tex}=\\frac{\\sqrt{3}}{4}{/tex}\xa0cos\xa070° [{tex}\\frac {1} {2}{/tex}\xa0+ cos 40°] [{tex}\\because{/tex}\xa0cos 60° =\xa0{tex}\\frac {1} {2}{/tex}]{tex}=\\frac{\\sqrt{3}}{4}{/tex}\xa0cos 70° +\xa0{tex}\\frac{\\sqrt{3}}{4}{/tex}\xa0cos 70° cos 40°{tex}=\\frac{\\sqrt{3}}{4}{/tex}\xa0cos 70° +\xa0{tex}\\frac{\\sqrt{3}}{8}{/tex}\xa0(2 cos 70° cos 40°){tex}= \\frac{\\sqrt{3}}{8}{/tex}\xa0[cos 70° + cos (70° + 40°) + cos (70° - 40°)] [from (i)]{tex}= \\frac{\\sqrt{3}}{8}{/tex}\xa0[cos 70° + cos 110° + cos 30°]{tex}= \\frac{\\sqrt{3}}{8}{/tex}\xa0[cos 70° + cos (180° - 70° +\xa0{tex}\\frac{\\sqrt{3}}{2}{/tex}]{tex}= \\frac{\\sqrt{3}}{8}{/tex}\xa0[cos 70° - cos 70° +\xa0{tex}\\frac{\\sqrt{3}}{2}{/tex}] [{tex}\\because{/tex}\xa0cos (180° - x) = - cos x]{tex}=\\frac{\\sqrt{3}}{8} \\times \\frac{\\sqrt{3}}{2}=\\frac{3}{16}{/tex}= RHS


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