Math Problem Statement

Prove that [tan (A-B /2)]/[tan (A + B / 2)] = (a-b)/(a+b)

Solution

To prove the identity:

tan(AB2)tan(A+B2)=aba+b,\frac{\tan\left(\frac{A - B}{2}\right)}{\tan\left(\frac{A + B}{2}\right)} = \frac{a - b}{a + b},

we use the tangent half-angle formulas and properties of trigonometric functions. Let's proceed step by step:

Step 1: Write tangent in terms of sine and cosine

The tangent of an angle can be expressed as: tanx=sinxcosx.\tan x = \frac{\sin x}{\cos x}.

Thus: tan(AB2)tan(A+B2)=sin(AB2)cos(AB2)sin(A+B2)cos(A+B2).\frac{\tan\left(\frac{A - B}{2}\right)}{\tan\left(\frac{A + B}{2}\right)} = \frac{\frac{\sin\left(\frac{A - B}{2}\right)}{\cos\left(\frac{A - B}{2}\right)}}{\frac{\sin\left(\frac{A + B}{2}\right)}{\cos\left(\frac{A + B}{2}\right)}}.

Simplify: tan(AB2)tan(A+B2)=sin(AB2)cos(A+B2)cos(AB2)sin(A+B2).\frac{\tan\left(\frac{A - B}{2}\right)}{\tan\left(\frac{A + B}{2}\right)} = \frac{\sin\left(\frac{A - B}{2}\right) \cos\left(\frac{A + B}{2}\right)}{\cos\left(\frac{A - B}{2}\right) \sin\left(\frac{A + B}{2}\right)}.

This becomes: tan(AB2)tan(A+B2)=sin(AB2)cos(A+B2)cos(AB2)sin(A+B2)cos(AB2)sin(A+B2).\frac{\tan\left(\frac{A - B}{2}\right)}{\tan\left(\frac{A + B}{2}\right)} = \frac{\sin\left(\frac{A - B}{2}\right) \cos\left(\frac{A + B}{2}\right) - \cos\left(\frac{A - B}{2}\right) \sin\left(\frac{A + B}{2}\right)}{\cos\left(\frac{A - B}{2}\right) \sin\left(\frac{A + B}{2}\right)}.

Step 2: Simplify using the sine subtraction formula

The numerator is of the form: sin(AB2)cos(A+B2)cos(AB2)sin(A+B2).\sin\left(\frac{A - B}{2}\right)\cos\left(\frac{A + B}{2}\right) - \cos\left(\frac{A - B}{2}\right)\sin\left(\frac{A + B}{2}\right).

This is equivalent to: sin(AB2A+B2).\sin\left(\frac{A - B}{2} - \frac{A + B}{2}\right).

Simplify the argument: AB2A+B2=B2.\frac{A - B}{2} - \frac{A + B}{2} = -\frac{B}{2}.

Thus, the numerator becomes: sin(B2).\sin\left(-\frac{B}{2}\right).

Using the odd property of sine (sin(x)=sin(x)\sin(-x) = -\sin(x)): sin(B2)=sin(B2).\sin\left(-\frac{B}{2}\right) = -\sin\left(\frac{B}{2}\right).

Step 3: Simplify the denominator

The denominator remains: cos(AB2)sin(A+B2).\cos\left(\frac{A - B}{2}\right)\sin\left(\frac{A + B}{2}\right).

Step 4: Combine the results

Now the expression is: tan(AB2)tan(A+B2)=sin(B2)cos(AB2)sin(A+B2).\frac{\tan\left(\frac{A - B}{2}\right)}{\tan\left(\frac{A + B}{2}\right)} = \frac{-\sin\left(\frac{B}{2}\right)}{\cos\left(\frac{A - B}{2}\right)\sin\left(\frac{A + B}{2}\right)}.

Use the identities and context to simplify further based on a=tan(A/2)a = \tan(A/2) and b=tan(B/2)b = \tan(B/2). Would you like to continue with this interpretation?

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Tangent Function
Half-Angle Formulas
Sine and Cosine

Formulas

tan(x) = sin(x)/cos(x)
sin(A - B) = sin(A)cos(B) - cos(A)sin(B)
sin(-x) = -sin(x)

Theorems

Tangent Half-Angle Formula
Sine and Cosine Identities

Suitable Grade Level

Grades 11-12