sin 8A/sin 4A - cos 8A/cos 4A = sec 4A | Prove the trigonometric identity | SciPiPupil

Question: Prove the following trigonometric identity: $\dfrac{sin 8A}{sin 4A} - \dfrac{cos 8A}{cos 4A}$ = sec 4A.

Solution:

LHS

$\dfrac{sin 8A}{sin 4A} - \dfrac{cos 8A}{cos 4A}$

$\dfrac{sin 8A.cos4A - cos8A.sin4A}{sin 4A.cos 4A}$

$\dfrac{sin (8A -4A)}{sin 4A.cos4A}$

$\dfrac{sin 4A}{sin4A.cos4A}$

$\dfrac{1}{cos 4A}$

$sec 4A$

= RHS

sin 8A/sin 4A - cos 8A/cos 4A = sec 4A | Prove the trigonometric identity | SciPiPupil