Question

tan5A- tan3A÷tan 5A+tan 3A = sin 2A÷sin 8A​

Answer 1

Answer:

Hey mate here is ur answer

Answer 2

Step-by-step explanation:

We know that

$tan \: A = \frac{sin \: A}{cos \: A} \\ \\ tan \: 5A = \frac{sin \: 5A}{cos \: 5A}$

$\frac{tan \: 5A - tan \: 3A}{tan \: 5A + tan \: 3A} \\ \\ = \frac{ \frac{sin \: 5A}{cos \: 5A} - \frac{sin \: 3A}{cos \: 3A}}{\frac{sin \: 5A}{cos \: 5A} + \frac{sin \: 3A}{cos \: 3A}} \\ \\ take \: LCM \\ \\ \frac{ \frac{sin \: 5A \: cos \: 3A - sin \: 3A \: cos \: 5A}{cos \: 3A \: cos \: 5A} }{\frac{sin \: 5A \: cos \: 3A + sin \: 3A \: cos \: 5A}{cos \: 3A \: cos \: 5A}} \\ \\ = \frac{{sin \: 5A \: cos \: 3A - sin \: 3A \: cos \: 5A} \: }{sin \: 5A \: cos \: 3A + sin \: 3A \: cos \: 5A}$

We know that

$sin(A + B) = sin \: A \: cos \: B + cos \: A \: sin \: B \\ \\ sin(A - B) = sin \: A \: cos \: B - cos \: A \: sin \: B$

apply these formulas

$= \frac{sin(5A - 3A)}{sin \: (5A + 3A)} \\ \\ = \frac{sin \: 2A}{sin \: 8A}$

Hence proved