Question

Find the value of a square +b square if a cos theta+b sin theta=4 and a sin
theta-b cos theta =3

Answer 1

Answer:

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Answer 2

Answer:

A cos theta + B sin theta = 4

[A cos theta + B sin theta]^2 = 16 ……….i

A sin theta - B cos theta = 3

[A sin theta - B cos theta]^2 = 9 ………….i i

Adding equation i and ii we get,

(A cos theta)^2 + (B sin theta)^2 + (A sin theta)^2 + (B cos theta)^2 + 2 AB cos theta sin theta - 2 AB cos theta sin theta = 25

A^2[(cos theta)^2 + (sin theta)^2] + B^2[(cos theta)^2 + (sin theta)^2] = 25

A^2 + B^2 = 25