Prove that sinA = k sinB, tan(1/2 (A-B)) = (k-1)/(k+1), and tan(1/2 (A+B)) = (1)/(cot(1/2 (A+B))).

Steps to Solve

1. Identify Given Information You have ( \tan A + \tan B = 5 ) and ( \tan A \tan B = \frac{1}{2} ).

2. Use the Tangent Addition Formula The tangent addition formula states: $$ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} $$

Substituting the given values: $$ \tan(A + B) = \frac{5}{1 - \frac{1}{2}} $$

3. Simplify the Expression Simplifying the denominator: $$ 1 - \frac{1}{2} = \frac{1}{2} $$

So, we have: $$ \tan(A + B) = \frac{5}{\frac{1}{2}} = 5 \times 2 = 10 $$

4. Find Cotangent of ( (A + B) ) The cotangent is the reciprocal of the tangent: $$ \cot(A + B) = \frac{1}{\tan(A + B)} = \frac{1}{10} $$

5. Express Sine and Cosine Relations Using ( \sin A = k \sin B ), express in terms of sine and cosine: $$ \sin A = k \sin B $$

So, $$ \frac{\sin A}{\sin B} = k $$

Using the sine difference formula: $$ \sin(A - B) = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) $$

6. Set Up Proportions Using K Substituting into the previous relation leads to: $$ \frac{\sin A + \sin B}{\sin A - \sin B} = \frac{k+1}{k-1} $$

Using the compound angle formulas, express:

• ( \sin \left( \frac{A+B}{2} \right) ) and ( \cos \left( \frac{A-B}{2} \right) ).

7. Final Relations with Tangent Combine identified values and finalize the relation. Using: $$ \tan\left(\frac{A+B}{2}\right) \cdot \cot\left(\frac{A-B}{2}\right) $$

8. Conclusion Arrive at the final relationship utilizing ( k ) values and verify.