# If 7 sin A + 24 cos A = 25, find the value of tan A.

Given : 7 sin A + 24 cos A = 25

Squaring both the sides, we get

(7 sin A + 24 cos A)2 = 625

49 sin2 A +576 cos2 A + 2(7sin A) (24cos A) = 625 [ (a + b)2 =a2 +b2 +2ab]

49 sin2 A +576 cos2 A + 336 cosA sinA = 625

Divide by cos2 θ, we get

49tan2 A +576+ 336 tanA = 625sec2 A

49tan2 A +576+ 336 tanA = 625(1 + tan2 A) [ 1+ tan2θ = sec2 θ]

49tan2 A +576+ 336 tanθA = 625+625 tan2 A

576tan2 A – 336tanA + 49 = 0

576tan2 A – 168 tanA – 168 tanA +49 = 0

24tanθ (24tan A 7) 7(24tan A 7) = 0

(24tan A – 7)2 = 0

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