Question

tan5A-tan3A/tan5A+tan3A = sin2A/sin8A

Answer 1

Answer:

$\frac{tan5A-tan3A}{tan5A+tan3A}=\frac{sin2A}{sin8A}$

Step-by-step explanation:

Formula used:

sin(A+B) = sinA cosB + cosA sinB

sin(A-B) = sinA cosB - cosA sinB

$\frac{tan5A-tan3A}{tan5A+tan3A}$

$=\frac{\frac{sin5A}{cos5A}-\frac{sin3A}{cos3A}}{\frac{sin5A}{cos5A}+\frac{sin3A}{cos3A}}$

$=\frac{sin5A.cos3A-cos5A.sin3A}{sin5A.cos3A+cos5A.sin3A}$

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

$=\frac{sin2A}{sin8A}$

Answer 2

Answer:

Step-by-step explanation:

Formula used:

sin(A+B) = sinA cosB + cosA sinB

sin(A-B) = sinA cosB - cosA sinB