Question

Please someone help me... i need full answer ​

Answer 1

Answer:  see proof below

Step-by-step explanation:

Use the following Sum to Product Identities:

$\sin x-\sin y=2\cos\bigg(\dfrac{x+y}{2}\bigg)\sin\bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x-\cos y=2\cos\bigg(\dfrac{x+y}{2}\bigg)\cos\bigg(\dfrac{x-y}{2}\bigg)$

Proof LHS →  RHS

$\text{LHS:}\qquad \qquad \qquad \qquad \dfrac{\sin 2A+\sin 5A-\sin A}{\cos A+\cos 2A+\cos 5A}$

$\text{Regroup:}\qquad \qquad \qquad \dfrac{\sin 2A+(\sin 5A-\sin A)}{\cos 2A+(\cos 5A+\cos A)}$

$\text{Sum to Product:}\qquad \dfrac{\sin 2A+2\cos\bigg(\dfrac{5A+A}{2}\bigg)\sin\bigg(\dfrac{5A-A}{2}\bigg)}{\cos2A+\cos\bigg(\dfrac{5A+A}{2}\bigg)\cos\bigg(\dfrac{5A-A}{2}\bigg)}$

$\text{Simplify:}\qquad \qquad \qquad \dfrac{\sin 2A+2\cos 3A\sin 2A}{\cos 2A+2\cos 3A\cos 2A}$

$\text{Factor:}\qquad \qquad \qquad \dfrac{\sin 2A(1+2\cos 3A)}{\cos 2A(1+2\cos 3A)}$

$\text{Simplify:}\qquad \qquad \qquad \dfrac{\sin 2A}{\cos 2A}\\\\.\qquad \qquad \qquad \qquad =\tan 2A$

LHS = RHS:  tan 2A = tan 2A  $\checkmark$

Answer 2

Answer:

hope it helps

plz mark brainliest✌️✌️