If 4SINB=3SIN(2A+B) :
Prove that:7COT(A+B)=COTA

Step-by-step explanation:

Given the expression 4sinB = 3sin(2A+B), we are to show that the expression 7cot(A+B) = cotA

Starting with the expression

4sinB= 3sin(2A+B)

Let us re write angle B = (A + B) - A

and 2A + B = (A + B) + A

Substituting the derived expression back into the original expression ww will have;

4Sin{(A + B) - A } = 3Sin{(A + B)+ A}

From trigonometry identity;

Sin(D+E) = SinDcosE + CosDSinE

Sin(D-E) = SinDcosE - CosDSinE

Applying this in the expression above;

4{Sin(A+B)CosA - Cos(A+B)SinA} = 3{Sin(A+B)CosA + Cos(A+B)sinA}

Open the bracket

4Sin(A+B)CosA - 4Cos(A+B)SinA = 3Sin(A+B)CosA + 3Cos(A+B)sinA

Collecting like terms

4Sin(A+B)CosA - 3Sin(A+B)cosA = 3Cos(A+B)sinA + 4Cos(A+B)sinA

Sin(A+B)CosA = 7Cos(A+B)sinA

Divide both sides by sinA

Sin(A+B)CosA/sinA= 7Cos(A+B)sinA/sinA

Since cosA/sinA = cotA, the expression becomes;

Sin(A+B)cotA = 7Cos(A+B)

Finally, divide both sides of the resulting equation by sin(A+B)

Sin(A+B)cotA/sin(A+B) = 7Cos(A+B)/sin(A+B)

CotA = 7cot(A+B) Proved!

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step-by-step explanation:

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