Question

Please help me!!!!!!!!!!!!!​

Answer 1

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° = $\frac{\sqrt2}{2}$

Proof LHS → RHS

Given:                       $cot\ (22\frac{1}{2})^o-tan\ (22\frac{1}{2})^o$

Rewrite Fraction:     $cot\ (\frac{45}{2})^o-tan\ (\frac{45}{2})^o$

Half-Angle Identity:   $\dfrac{sin(45)^o}{1-cos(45)^o}-\dfrac{sin(45)^o}{1+cos(45)^o}$

Substitute:                  $\dfrac{\frac{\sqrt2}{2}}{1-\frac{\sqrt2}{2}}-\dfrac{\frac{\sqrt2}{2}}{1+\frac{\sqrt2}{2}}$

Simplify:                      $\dfrac{\frac{\sqrt2}{2}}{\frac{2-\sqrt2}{2}}-\dfrac{\frac{\sqrt2}{2}}{\frac{2+\sqrt2}{2}}$

                               $=\dfrac{\sqrt2}{2-\sqrt2}-\dfrac{\sqrt2}{2+\sqrt2}$

                               $=\dfrac{\sqrt2}{2-\sqrt2}\bigg(\dfrac{2+\sqrt2}{2+\sqrt2}\bigg)-\dfrac{\sqrt2}{2+\sqrt2}\bigg(\dfrac{2-\sqrt2}{2-\sqrt2}\bigg)$

                               $=\dfrac{2\sqrt2+2}{4-2}-\dfrac{2\sqrt2-2}{4-2}$

                               $=\dfrac{4}{2}$

                               = 2

LHS = RHS:  2 = 2  $\checkmark$