Answer: see both proofs below
Step-by-step explanation:
Use Difference Identity: tan (A - B) = (tan A - tan B)/(1 + tanA · tanB)
Use Unit Circle to evaluate: tan (π/4) = 1
Use Tangent Identity: tanA = (sinA)/(cosA)
Use Half-Angle Identities:
[tex]\cos \dfrac{A}{2}=\sqrt{\dfrac{1+\cos A}{2}}\\\\\\\sin \dfrac{A}{2}=\sqrt{\dfrac{1-\cos A}{2}}[/tex]
Part 1 Proof LHS → Middle
LHS: [tex]\tan\bigg(\dfrac{\pi}{4}-\dfrac{A}{2}\bigg)[/tex]
Difference Identity: [tex]\dfrac{\tan (\frac{\pi}{4})-\tan(\frac{A}{2})}{1+\tan(\frac{\pi}{4})\cdot \tan(\frac{A}{2})}[/tex]
Unit Circle: [tex]\dfrac{1-\tan(\frac{A}{2})}{1+ \tan(\frac{A}{2})}[/tex]
[tex]\text{Tangent Identity:}\qquad \qquad \dfrac{\frac{\cos\frac{A}{2}-\sin\frac{A}{2}}{\cos\frac{A}{2}}}{\frac{\cos\frac{A}{2}+\sin\frac{A}{2}}{\cos\frac{A}{2}}}[/tex]
Simplify: [tex]\dfrac{\cos\frac{A}{2}-\sin\frac{A}{2}}{\cos\frac{A}{2}+\sin\frac{A}{2}}[/tex]
LHS = Middle: [tex]\dfrac{\cos\frac{A}{2}-\sin\frac{A}{2}}{\cos\frac{A}{2}+\sin\frac{A}{2}}=\dfrac{\cos\frac{A}{2}-\sin\frac{A}{2}}{\cos\frac{A}{2}+\sin\frac{A}{2}}\qquad \checkmark[/tex]
Part 2 Proof Middle → RHS
Middle: [tex]\dfrac{\cos\frac{A}{2}-\sin\frac{A}{2}}{\cos\frac{A}{2}+\sin\frac{A}{2}}[/tex]
[tex]\text{Half-Angle Identity:}\qquad \qquad \dfrac{\sqrt{\frac{1+\cos A}{2}}-\sqrt{\frac{1-\cos A}{2}}}{\sqrt{\frac{1+\cos A}{2}}+\sqrt{\frac{1-\cos A}{2}}}[/tex]
Simplify: [tex]\dfrac{\sqrt{1+\cos A}-\sqrt{1-\cos A}}{\sqrt{1+\cos A}+\sqrt{1-\cos A}}[/tex]
Rationalize Denominator: [tex]\dfrac{\sqrt{1+\cos A}-\sqrt{1-\cos A}}{\sqrt{1+\cos A}+\sqrt{1-\cos A}}\bigg(\dfrac{\sqrt{1+\cos A}-\sqrt{1-\cos A}}{\sqrt{1+\cos A}-\sqrt{1-\cos A}}\bigg)[/tex]
[tex]=\dfrac{1+\cos A-2\sqrt{1-\cos^2 A}+1-\cos A}{1+\cos A-(1-\cos A)}[/tex]
Simplify: [tex]\dfrac{2-2\sqrt{1-\cos^2 A}}{2\cos A}[/tex]
[tex]=\dfrac{2-2\sqrt{sin^2 A}}{2\cos A}[/tex]
[tex]= \dfrac{2-2\sin A}{2\cos A}[/tex]
Factor: [tex]\dfrac{2(1-\sin A)}{2(\cos A)}[/tex]
Simplify: [tex]\dfrac{1-\sin A}{\cos A}[/tex]
Expand: [tex]\dfrac{1-\sin A}{\cos A}\bigg(\dfrac{1+\sin A}{1+\sin A}\bigg)[/tex]
[tex]=\dfrac{1-\sin^2 A}{\cos A(1+\sin A)}[/tex]
Simplify: [tex]\dfrac{\cos^2 A}{\cos A(1+\sin A)}[/tex]
[tex]=\dfrac{\cos A}{1+\sin A}[/tex]
Middle = RHS: [tex]\dfrac{\cos A}{1+\sin A}=\dfrac{\cos A}{1+\sin A}\qquad \checkmark[/tex]
