Prove that: cos2Acos3A-cos2Acos7A+cosAcos10A /sin4Asin3A-sin2Asin5A+sin4Asin7A =cot6Acot5A
Answers
your question is -> (cos2A.cos3A - cos2A.cos7A + cosA.cos10A)/(sin4A.sin3A - sin2A.sin5A + sin4A.sin7A) = cot6A.cot5A
proof : LHS = (cos2A.cos3A - cos2A.cos7A + cosA.cos10A)/(sin4A.sin3A - sin2A.sin5A + sin4A.sin7A)
using formula,
cosA.cosB = 1/2 [cos(A + B) + cos(A - B)]
cosA.cosB = 1/2 [cos(A + B) + cos(A - B)] sinA.sinB = 1/2 [cos(A - B) - cos(A + B)]
so, cos2A.cos7A = 1/2 [cos(2A + 7A) + cos(7A - 2A) ]
= 1/2 [cos9A + cos5A]
similarly, cos2A .cos3A = 1/2 [cos5A + cosA]
cosA.cos10A = 1/2 [cos11A + cos9A]
sin4A.sin3A = 1/2 [cos(4A - 3A) - cos(4A + 3A)
= 1/2[cosA - cos7A]
simiarly, sin2A.sin5A = 1/2[cos(5A - 2A) - cos(5A + 2A)] = 1/2[cos3A - cos7A]
sin4A.sin7A = 1/2[cos3A - cos11A]
then, [1/2(cos5A + cosA - cos9A - cos5A + cos11A + cos9A)]/[1/2(cosA- cos7A - cos3A + cos7A + cos3A - cos11A)]
= (cosA + cos11A)/(cosA - cos11A)
using formula,
cosC + cosD = 2cos(C + D)/2.cos(C - D)/2
cosC + cosD = 2cos(C + D)/2.cos(C - D)/2cosC - cosD = 2sin(C + D)/2.sin(D - C)/2
= {2cos(A+11A)/2.cos(11A-A)/2}/{2sin(A + 11A)/2.sin(11A-A)/2}
= {cos6A.cos5A}/{sin6A.sin5A}
= cot6A.cot5A = RHS
Answer:
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