Question

Please help me guys

a) If 2 Tan a = 3 Tan B, then show that Tan (a-B) =5 sin2b/5Cos 2B-1
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Answer 1

Given that :-

• 2 tan A = 3 tan B

Prove that :-

• Tan (a-B) = [ sin 2B/(5 – cos 2B)]

Taking RHS :-

= sin 2B/(5 – cos 2B)] = [(2 tan B)/(1 + tan2B)]/ [5 – {(1 – tan2B)/(1 + tan2B)}]

= (2 tan B)/ [5(1 + tan2B) – 1 + tan2B]

= (2 tan B)/[5 + 5 tan2B – 1 + tan2B)

= (2 tan B)/(4 + 6 tan2B)

= (2 tan B)/[2(2 + 3 tan2B)]

= tan B/(2 + 3 tan B tan B)

Dividing the numerator and denominator by 2,

= [(1/2) tan B]/ [1 + (3/2) tan B tan B]

= [(3/2) tan B – tan B]/[1 + tan A tan B] {from (i)}

= (tan A – tan B)/(1 + tan A tan B)

= tan(A – B) = LHS