Question

PROVE IT. Sin(4a-2B)+sin(4B-2A)/cos(4A-2B)+(cos(4B-2A)=tan(A+B)

Answer 1

Answer:   Proved.

   

Step-by-step explanation:  We are given to prove the following inequality-

$\dfrac{\sin(4A-2B)+\sin(4B-2A)}{\cos(4A-2B)+\cos(4B-2A)}=\tan(A+B).$

Since this is a trigonometric problem involving angles A and B of a right-angled triangle, we need the following relations from trigonometry

$\dfrac{\sin \theta}{\cos \theta}=\tan \theta,\\\\\sin \theta_1+\sin \theta_2=2\sin\dfrac{\theta_1+\theta_2}{2}\cos\dfrac{\theta_1-\theta_2}{2}\\\\\textup{and}\\\\\cos \theta_1+\cos \theta_2=2\cos\dfrac{\theta_1+\theta_2}{2}\cos\dfrac{\theta_1-\theta_2}{2}.$

Now,

$L.H.S\\\\=\dfrac{\sin(4A-2B)+\sin(4B-2A)}{\cos(4A-2B)+\cos(4B-2A)}\\\\\\\\=\dfrac{2\sin(\dfrac{4A-2B+4B-2A}{2})\cos(\dfrac{4A-2B-4B+2A}{2})}{2\cos(\dfrac{4A-2B+4B-2A}{2})\cos(\dfrac{4A-2B-4B+2A}{2})}\\\\\\\\=\dfrac{\sin(A+B)\cos(3A-3B)}{\cos(A+B)\cos(3A-3B)}\\\\=\tan(A+B)\\\\=R.H.S.$

Hence proved.