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Please solve this question 2sin(square)B+4cos(A+B)sinAcosB+cos2(A+B)=cos2A

Please solve this question 2sin(square)B+4cos(A+B)sinAcosB+cos2(A+B)=cos2A

Grade:11

1 Answers

Himanshu
103 Points
7 years ago
We can write it as :-
1 – cos2B + 4.cos(A+B).sinA.cosB + 2.cos2(A+B) – 1 = cos2A
cos2A + cos2B = 4.cos(A+B).sinA.cosB + 2cos2(A+B)
2.cos(A+B).cos(A-B) – 4.cos(A+B).sinA.cosB = 2cos2(A+B)
cos(A+B).cos(A-B) – 2.cos(A+B).sinA.cosB – cos2(A+B) = 0
cos(A+B) { cos(A-B) – 2.sinA.cosB – cos(A+B) } = 0
cos(A+B) { cosA.cosB + sinA.sinB – cosA.cosB + sinA.sinB – 2.sinA.cosB } = 0
cos(A+B) [ sinA {sinB – 2.cosB} ] = 0
Now you can easily solve this three equations individually and merge them in the answer.
cos(A+B) = 0          ...i)
sinA = 0                 …ii)
sinB – 2cosB = 0    ...iii)
Solve them and you will get the required answer.

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