if tanB = (nsinAcosA) /(1 - ncos^2A) then tan(A + B) = ?
iamKapilFan:
sorry for such comment but tomorrow is my physics exam nd I was studying from the morning so now I don't want to study at all
so sorry
I can't answer your question
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Oh thanks
nd sorry again
it's k
yup...
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tanB= nsinA cos A)/(1- n cos^2A)
= 2n sinA cosA/2(1- n cos^2A)
= n Sin2A/2(1- n cos^2A)
= n 2 tanA/2(1+tan^2A)( 1- n cos^2A)
= n tan A/( 1 - n cos^2A + tan^2A- n sin^2A)
= ntanA/( 1-n( Cos^2A+ Sin^2A) + tan^2A)
= n tan A/( 1 - n + tan^2A)
tan( A+ B) = tan A + tan B)/( 1- tan A tan B)
= tan A + n tanA/( 1- n + tan^2A))/( 1 - tanA( n tan A/( 1- n + tan^2 A))
= tanA - n tanA + tan^3 A + n tanA)/( 1- n + tan^2 A - n tan^2 A)
= tan A + tan^3A)/ ( 1- n + tan^2A( 1- n))
= tanA + tan^3 A)/( 1-n)( 1+ tan^2 A)
= tanA( 1+ tan^2A)/( 1-n)( 1+ tan^2A)
= TanA/( 1- n)
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