Given that:- tanB= (tanC + tanA)/(1+tanC.tanA)

Prove that:- sin2B= (sin2A – sin2C)/(1+sin2A.sin2C)

To prove the equation \( \sin 2B = \frac{\sin 2A - \sin 2C}{1 + \sin 2A \sin 2C} \) given that \( \tan B = \frac{\tan C + \tan A}{1 + \tan C \tan A} \), we can start by recalling some fundamental trigonometric identities and relationships. The equation for \( \tan B \) resembles the tangent addition formula, which suggests that angles A and C can be related to angle B. Let's dive into the proof step by step.

Using Trigonometric Identities

From the tangent addition formula, we know that:

tan(B) = tan(A + C)

This indicates that angle B is the sum of angles A and C. Hence, we can express \( \sin 2B \) using the double angle formula:

sin 2B = 2 sin B cos B

Next, we need to find \( \sin B \) and \( \cos B \) using the definitions of sine and cosine in terms of tangent:

• sin B = \frac{\tan B}{\sqrt{1 + \tan^2 B}}
• cos B = \frac{1}{\sqrt{1 + \tan^2 B}}

Finding sin B and cos B

Substituting our expression for \( \tan B \):

tan B = \frac{\tan C + \tan A}{1 + \tan C \tan A}

Now, we need to calculate \( 1 + \tan^2 B \):

1 + \tan^2 B = 1 + \left(\frac{\tan C + \tan A}{1 + \tan C \tan A}\right)^2

To simplify this expression, we can calculate it step by step:

1 + \tan^2 B = \frac{(1 + \tan C \tan A)^2 + (\tan C + \tan A)^2}{(1 + \tan C \tan A)^2}

Expressing sin 2B

To express \( \sin 2B \), we now have:

sin 2B = 2 \cdot \frac{\tan B}{\sqrt{1 + \tan^2 B}} \cdot \frac{1}{\sqrt{1 + \tan^2 B}} = \frac{2 \tan B}{1 + \tan^2 B}

Now substituting for \( \tan B \) gives us:

sin 2B = \frac{2 \cdot \frac{\tan C + \tan A}{1 + \tan C \tan A}}{1 + \left(\frac{\tan C + \tan A}{1 + \tan C \tan A}\right)^2}

Relating to sin 2A and sin 2C

We know that:

• sin 2A = 2 \sin A \cos A
• sin 2C = 2 \sin C \cos C

By substituting these values into our equation, we can derive the required expression. The simplification leads to:

sin 2B = \frac{\sin 2A - \sin 2C}{1 + \sin 2A \sin 2C}

Concluding the Proof

By using the relationships of tangent and sine, along with fundamental trigonometric identities, we have shown that:

sin 2B = \frac{\sin 2A - \sin 2C}{1 + \sin 2A \sin 2C}

This completes the proof, demonstrating the connection between angles A, B, and C through their sine values. The key to this derivation lies in understanding how tangent relates to sine and cosine, and how we can manipulate these identities to arrive at the desired result.