Question

tan(a-b) = sin2b/3-cos2b​

Answer 1

answer

just break the formulas and convert them into tan and do some manipulations over there. Then get your answer. I've given you a picture of the calculation. just go through it.

Answer 2

Answer:

$tan(a)=2tan(b)$

Step-by-step explanation:

RHS(simplification by using elementary indentities:

$=\frac{sin2b}{3-cos2b} \\= \frac{2sinbcosb}{2+(1-cos2b)} \\= \frac{2sinbcosb}{2+2sin^2b} \\=\frac{sinbcosb}{1+sin^2b} \\.$

LHS:

$tan(a-b)=\frac{tan(a)-tan(b)}{1+tan(a)tan(b)} .$

Solving we get

$\implies tan(a)[1+sin^2(b)]-tan(b)[1+sin^2(b)]=sin(b)cos(b)+tan(a)tan(b)sin(b)cos(b)\\\implies tan(a)-tan(b)+tan(a)sin^2(b)-\frac{sin^3(b)}{cos(b)} = sin(b)cos(b)+tan(a)sin^2(b)\\\implies tan(a)cos(b)-sin(b)-sin^3(b)=sin(b)cos^2(b)=sin(b)-sin^3(b)\\\implies tan(a)=2tan(b)$