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ollegr [7]
3 years ago
13

Help me please!!!!!!​

Mathematics
1 answer:
Vaselesa [24]3 years ago
7 0

Answer:  see both proofs below

<u>Step-by-step explanation:</u>

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:

\cos \dfrac{A}{2}=\sqrt{\dfrac{1+\cos A}{2}}\\\\\\\sin \dfrac{A}{2}=\sqrt{\dfrac{1-\cos A}{2}}

<u>Part 1   Proof LHS → Middle</u>

LHS:                                \tan\bigg(\dfrac{\pi}{4}-\dfrac{A}{2}\bigg)

Difference Identity:        \dfrac{\tan (\frac{\pi}{4})-\tan(\frac{A}{2})}{1+\tan(\frac{\pi}{4})\cdot \tan(\frac{A}{2})}

Unit Circle:                      \dfrac{1-\tan(\frac{A}{2})}{1+ \tan(\frac{A}{2})}

\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}}}

Simplify:                             \dfrac{\cos\frac{A}{2}-\sin\frac{A}{2}}{\cos\frac{A}{2}+\sin\frac{A}{2}}

LHS = Middle:  \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

<u>Part 2   Proof  Middle → RHS</u>

Middle:                                     \dfrac{\cos\frac{A}{2}-\sin\frac{A}{2}}{\cos\frac{A}{2}+\sin\frac{A}{2}}

\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}}}

Simplify:                               \dfrac{\sqrt{1+\cos A}-\sqrt{1-\cos A}}{\sqrt{1+\cos A}+\sqrt{1-\cos A}}

Rationalize Denominator:  \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)

                                          =\dfrac{1+\cos A-2\sqrt{1-\cos^2 A}+1-\cos A}{1+\cos A-(1-\cos A)}

Simplify:                                \dfrac{2-2\sqrt{1-\cos^2 A}}{2\cos A}

                                         =\dfrac{2-2\sqrt{sin^2 A}}{2\cos A}

                                         = \dfrac{2-2\sin A}{2\cos A}

Factor:                                   \dfrac{2(1-\sin A)}{2(\cos A)}

Simplify:                               \dfrac{1-\sin A}{\cos A}

Expand:                                \dfrac{1-\sin A}{\cos A}\bigg(\dfrac{1+\sin A}{1+\sin A}\bigg)

                                          =\dfrac{1-\sin^2 A}{\cos A(1+\sin A)}

Simplify:                                 \dfrac{\cos^2 A}{\cos A(1+\sin A)}

                                          =\dfrac{\cos A}{1+\sin A}      

Middle = RHS:   \dfrac{\cos A}{1+\sin A}=\dfrac{\cos A}{1+\sin A}\qquad \checkmark            

                       

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