Question

prove that : tan5A+tan3A / tan5A-tan3A = 4cos2Acos4A

Answer 1

L.H.S=R.H.S bro very nice question...........
I love math's....
join me.....

Answer 2

Answer:

$\frac{tan5A+tan3A}{tan5A-tan3A}=4cos2Acos4A$

Step-by-step explanation:

$LHS=\frac{tan5A+tan3A}{tan5A-tan3A}\\=\frac{\frac{sin5A}{cos5A}+\frac{sin3A}{cos3A}}{\frac{sin5A}{cos5A}-\frac{sin3A}{cos3A}}\\=\frac{\frac{sin5Acos3A+sin3Acos5A}{cos5Acos3A}}{\frac{sin5Acos3A-sin3Acos5A}{cos5Acos3A}}\\=\frac{sin5Acos3A+sin3Acos5A}{sin5Acos3A-sin3Acos5A}\\=\frac{sin(5A+3A)}{sin(5A-3A)}\\=\frac{sin8A}{sin2A}\\=\frac{2sin4Acos4A}{sin2A}\\=\frac{2\times 2sin2Acos2Acos4A}{sin2A}\\=4cos2Acos4A\\=RHS$

Therefore,

$\frac{tan5A+tan3A}{tan5A-tan3A}=4cos2Acos4A$

•••♪