Question

please answer this please fast ​

Answer 1

Answer:

A+2B = π/2

Step-by-step explanation:

Given, 3^2cos(A)+2cos^2(B)

⇒2cos^2(B)−1=4−3cos

⇒cos2(B)=3(1−cos^2(A))=3sin^2(A)   ..... (1)

and 2cosBsinB=3sinAcosA

sin2B=3sinAcosA     .... (2)

Now, cos(A+2B)=cosAcos2B−sinAsin2B

=cosA(3sin^2(A)-sinA(3sinAcosA)=0     ....[using eqs. (1) and (2)]

⇒A+2B = π/2

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