Question

Help please!

Answer 1

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:   $\tan A=\dfrac{\sin A}{\cos A}$

Proof LHS → RHS:

$\text{LHS:}\qquad \qquad \dfrac{\cos \bigg(\dfrac{\pi}{2}+x\bigg)}{\cos \bigg(\pi +x\bigg)}$

$\text{Sum Difference:}\qquad \dfrac{\cos \dfrac{\pi}{2}\cdot \cos x-\sin \dfrac{\pi}{2}\cdot \sin x}{\cos \pi \cdot \cos x-\sin \pi \cdot \sin x}$

$\text{Unit Circle:}\qquad \qquad \dfrac{0\cdot \cos x-1\cdot \sin x}{-1\cdot \cos x-0\cdot \sin x}$

                            $=\dfrac{-\sin x}{-\cos x}$

Quotient:                 tan x

LHS = RHS $\checkmark$

Answer 2

Answer:

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: \tan A=\dfrac{\sin A}{\cos A}tanA=

cosA

sinA

Proof LHS → RHS:

\text{LHS:}\qquad \qquad \dfrac{\cos \bigg(\dfrac{\pi}{2}+x\bigg)}{\cos \bigg(\pi +x\bigg)}LHS:

cos(π+x)

cos(

2

π

+x)

\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}Sum Difference:

cosπ⋅cosx−sinπ⋅sinx

cos

2

π

⋅cosx−sin

2

π

⋅sinx

\text{Unit Circle:}\qquad \qquad \dfrac{0\cdot \cos x-1\cdot \sin x}{-1\cdot \cos x-0\cdot \sin x}Unit Circle:

−1⋅cosx−0⋅sinx

0⋅cosx−1⋅sinx

=\dfrac{-\sin x}{-\cos x}=

−cosx

−sinx

Quotient: tan x

LHS = RHS \checkmark✓