Question

If A=π\3 and B=π/6 then prove that Tan (A-B) = TanA-TanB/1-TanATanB​

Answer 1

Step-by-step explanation:

tan (A+B) = [tan A + tan B]/[1 - tan A tan B]

RHS = [tan A + tan B]/[1 - tan A tan B]

=[(sin A/cos A) + (sin B/cos B)]/[1-(sin A/cos A)(sin B/cos B)

= [sin A cos B + cos A sin B]/[cos A cos B][1 - sin A sin B/(cos A cos B)]

= sin (A+B)/{[cos A cos B][cos A cos B - sin A sin B]/(cos A cos B)}

= sin (A+B)/[cos A cos B - sin A sin B

= sin (A+B)/cos (A+B)

= tan (A+B) = LHS.