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Ex 3.3, 21 Prove that (cos⁡4𝑥 + cos⁡3𝑥 + cos⁡2𝑥)/(sin⁡4𝑥 + sin⁡3𝑥 + sin⁡2𝑥 ) = cot 3x Solving L.H.S Solving Numerator and Denominator separately We know that cos x + cos y = 2cos ((𝑥 + 𝑦)/2) cos ((𝑥 −𝑦)/2) Replacing x by 4x and y by 2x cos 4x + cos 2x = 2cos ((4𝑥 + 2𝑥)/2). cos ((4𝑥 − 2𝑥)/2) = 2cos (6𝑥/2). cos (2𝑥/2) = 2 cos 3x . cos x Now cos 4x + cos 2x + cos 3x = 2cos 3x . cos x + cos 3x = cos 3x (2cos x + 1) Similarly, Solving denominator sin 4x + sin 2x + sin 3x We know that sin x + sin y = 2sin ((𝑥 + 𝑦)/2) sin ((𝑥 −𝑦)/2) Replacing x by 4x and y by 2x sin 4x + sin 2x = 2 sin ((4𝑥 +2𝑥)/2). cos ((4𝑥 − 2𝑥)/2) = 2 sin (6𝑥/2). cos (2𝑥/2) = 2 sin 3x . cos x Now, sin 4x + sin 2x + sin 3x = sin 4x + sin 2x + sin 3x = 2sin 3x . cos x + sin 3x = sin 3x (2cos x + 1) Hence, our equation becomes (cos⁡4𝑥 + cos⁡3𝑥 + cos⁡2𝑥)/(sin⁡4𝑥 + sin⁡3𝑥 + sin⁡2𝑥 ) = (cos⁡3𝑥 (2 𝑐𝑜𝑠 𝑥 +1 ))/(sin⁡3𝑥 (2 𝑐𝑜𝑠 𝑥 +1 )) = cos⁡3𝑥/sin⁡3𝑥 = cot 3x = R.H.S. Hence R.H.S. = L.H.S. Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.