If tan B = mSinACosA/ 1-mSin^2A , prove that tan(A-B)=(1-m)tanA

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Ravneet · Answer

ANALYSE THE QUESTION : tan B = mSinACosA/ 1-mSin^2A we have to find tan(A-B) which is equal to (tanA – tanB )/(1+tanAtanB) ------------(1) IN THIS QUESTION, JUST PUT THE VALUE OF tanB in upper equation (1) after solving, you will get [tanA(1-msin 2 A) – msinAcos A ] / [1- msin 2 A + tanAmsinAcosA ] JUST LOOK AT THE DENOMINATOR, IT IS BECOMING 1 ( write tanA = sinA/cosA AND msin 2 A term will be cancelled ) NOW YOU ARE LEFT WITH NUMERATOR, mutiply and divide msinAcosA by cosA AND take tanA common. tanA [1-m(sin 2 A + cos 2 A ) tanA [ 1-m(1) ]

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If tan B = mSinACosA/ 1-mSin^2A , prove that tan(A-B)=(1-m)tanA

If tan B = mSinACosA/ 1-mSin^2A , prove that tan(A-B)=(1-m)tanA

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