Answer: see proof below
Step-by-step explanation:
Use the following Half-Angle Identities: tan (A/2) = (sinA)/(1 + cosA)
cot (A/2) = (sinA)/(1 - cosA)
Use the Pythagorean Identity: cos²A + sin²B = 1
Use Unit Circle to evaluate: cos 45° = sin 45° = [tex]\frac{\sqrt2}{2}[/tex]
Proof LHS → RHS
Given: [tex]cot\ (22\frac{1}{2})^o-tan\ (22\frac{1}{2})^o[/tex]
Rewrite Fraction: [tex]cot\ (\frac{45}{2})^o-tan\ (\frac{45}{2})^o[/tex]
Half-Angle Identity: [tex]\dfrac{sin(45)^o}{1-cos(45)^o}-\dfrac{sin(45)^o}{1+cos(45)^o}[/tex]
Substitute: [tex]\dfrac{\frac{\sqrt2}{2}}{1-\frac{\sqrt2}{2}}-\dfrac{\frac{\sqrt2}{2}}{1+\frac{\sqrt2}{2}}[/tex]
Simplify: [tex]\dfrac{\frac{\sqrt2}{2}}{\frac{2-\sqrt2}{2}}-\dfrac{\frac{\sqrt2}{2}}{\frac{2+\sqrt2}{2}}[/tex]
[tex]=\dfrac{\sqrt2}{2-\sqrt2}-\dfrac{\sqrt2}{2+\sqrt2}[/tex]
[tex]=\dfrac{\sqrt2}{2-\sqrt2}\bigg(\dfrac{2+\sqrt2}{2+\sqrt2}\bigg)-\dfrac{\sqrt2}{2+\sqrt2}\bigg(\dfrac{2-\sqrt2}{2-\sqrt2}\bigg)[/tex]
[tex]=\dfrac{2\sqrt2+2}{4-2}-\dfrac{2\sqrt2-2}{4-2}[/tex]
[tex]=\dfrac{4}{2}[/tex]
= 2
LHS = RHS: 2 = 2 [tex]\checkmark[/tex]
