Ravneet
Last Activity: 6 Years ago
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-msin2A) – msinAcosA ] / [1- msin2A + tanAmsinAcosA ]
JUST LOOK AT THE DENOMINATOR, IT IS BECOMING 1 (write tanA = sinA/cosA AND msin2A term will be cancelled )
NOW YOU ARE LEFT WITH NUMERATOR,
mutiply and divide msinAcosA by cosA AND take tanA common.
tanA [1-m(sin2A + cos2A )
tanA [ 1-m(1) ]