Answer: see proof below
Step-by-step explanation:
Use the Sum & Difference Identity: cos (A + B) = cos A · cos B - sin A · sin B
Recall the following from Unit Circle: cos (π/2) = 0, sin (π/2) = 1
cos (π) = -1, sin (π) = 0
Use the Quotient Identity: [tex]\tan A=\dfrac{\sin A}{\cos A}[/tex]
Proof LHS → RHS:
[tex]\text{LHS:}\qquad \qquad \dfrac{\cos \bigg(\dfrac{\pi}{2}+x\bigg)}{\cos \bigg(\pi +x\bigg)}[/tex]
[tex]\text{Sum Difference:}\qquad \dfrac{\cos \dfrac{\pi}{2}\cdot \cos x-\sin \dfrac{\pi}{2}\cdot \sin x}{\cos \pi \cdot \cos x-\sin \pi \cdot \sin x}[/tex]
[tex]\text{Unit Circle:}\qquad \qquad \dfrac{0\cdot \cos x-1\cdot \sin x}{-1\cdot \cos x-0\cdot \sin x}[/tex]
[tex]=\dfrac{-\sin x}{-\cos x}[/tex]
Quotient: tan x
LHS = RHS [tex]\checkmark[/tex]
