secA-tanA = K, then sinA = ? sin x + sin y = a ; cos x + cos y = b, then tan(x+y/2) = ? (ans in terms of a and b)

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Subhiksha · Answer

secA- Tan A =K 1-sinA/ ( 1- sin 2 A) ½ =k k 2 = 1- sin A / 1+ sin A By applying componendo and dividendo, sinA = 1- k 2 /1+k 2 sinx + siny= 2cos(x-y/2)sin(x+y/2) =a Cosx+ Cosy= 2cos(x-y/2)cos(x+y/2) = b On dividing both the above equations, = a/b tan(x+y/2)

Subhiksha · Answer

rewrite the last equationasTan (

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secA-tanA = K, then sinA = ? sin x + sin y = a ; cos x + cos y = b, then tan(x+y/2) = ? (ans in terms of a and b)

_{secA-tanA = K, then sinA = ?}

_{sin x + sin y = a ; cos x + cos y = b, then tan(x+y/2) = ? (ans in terms of a and b)}

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secA-tanA = K, then sinA = ? sin x + sin y = a ; cos x + cos y = b, then tan(x+y/2) = ? (ans in terms of a and b)

secA-tanA = K, then sinA = ? sin x + sin y = a ; cos x + cos y = b, then tan(x+y/2) = ? (ans in terms of a and b)

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_{secA-tanA = K, then sinA = ?}

_{sin x + sin y = a ; cos x + cos y = b, then tan(x+y/2) = ? (ans in terms of a and b)}