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Home / Doubts and Solutions/JEE FOUNDATION/Class 10/Mathematics

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Asked by nandrammeena2 20th February 2019, 10:08 PM
Answered by Expert
Answer:
3 sin2A + 2 sin2B = 0 , hence  sin2B = (-3/2)sin2A ....................(1)
 
3 sin2A + 2 sin2B = 1    or   3 sin2A = 1 - 2 sin2B = cos 2B  ......................(2)
 
cos(A+2B) = cosA cos2B + sinA sin2B  ..........................(3)
 
let us substitute for sin2B and cos2B in eqn.(3),   using eqn.(1) and eqn.(2)
 
cos(A+2B) = cosA (3sin2A ) +(-3/2) sinA sin2A   = cosA (3sin2A ) +(-3/2) sinA (2 sinA cosA )  = 0
 
Hence (A+2B) = π/2
Answered by Expert 21st February 2019, 12:41 AM
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