Math, asked by devanshchoudhary17mc, 11 months ago

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Answered by Anonymous
2

2sin5A Cos2A = Sin7A + Sin3A

2sin6AcosA = sin 7A + sin 5A

2SinA Sin2A = CosA- cos 3A

2Cos2A cos 3A =Cos5A + Cos A

Multiply by 2 in N AND D

sin7 A + Sin 3A - sin7A - sin5A)/( cosA- cos 3A - cos5A - cosA)

= sin3A- sin5A)/( cos 3A- cos5A)

= - 2sin Acos 4A )/( -2 cos A cos4A)

= tanA

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Answered by ItzMissRoyalPriyanka
8

Answer:

2sin5A Cos2A = Sin7A + Sin3A

2sin6AcosA = sin 7A + sin 5A

2SinA Sin2A = CosA- cos 3A

2Cos2A cos 3A -Cos5A + Cos A

Multiply by 2 in N AND D

sin7 A+ Sin 3A-sin7A - sin5A)/(cosA cos 3A - cos5A - cosA)

= sin3A-sin5A)/(cos 3A- cos5A)

= -2sin Acos 4A)/(-2 cos A cos4A)

= tanA

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