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LekaFEV [45]
3 years ago
12

Please help me!!!!!​

Mathematics
1 answer:
Bas_tet [7]3 years ago
4 0

Answer:  see proof below

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

Use the Sum & Difference Identity: tan (A - B) = (tanA - tanB)/(1 + tanA tanB)

Use the Half-Angle Identity: tan (A/2) = (1 - cosA)/(sinA)

Use the Unit Circle to evaluate tan (π/4) = 1

Use Pythagorean Identity:     cos²A + sin²A = 1

<u>Proof LHS → RHS</u>

\text{Given:}\qquad \qquad \qquad\dfrac{2\tan\bigg(\dfrac{\pi}{4}-\dfrac{A}{2}\bigg)}{1+\tan^2\bigg(\dfrac{\pi}{4}-\dfrac{A}{2}\bigg)}

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

\text{Substitute:}\qquad \qquad \dfrac{2 \bigg( \frac{1-\tan\frac{A}{2}}{1+\tan\frac{A}{2}}\bigg)}{1+ \bigg( \frac{1-\tan\frac{A}{2}}{1+\tan\frac{A}{2}}\bigg)^2}

\text{Simplify:}\qquad \qquad \qquad \dfrac{1-\tan^2\frac{A}{2}}{1+\tan^2\frac{A}{2}}

\text{Half-Angle Identity:}\qquad \quad \dfrac{1-(\frac{1-\cos A}{\sin A})^2}{1+(\frac{1-\cos A}{\sin A})^2}

\text{Simplify:}\qquad \qquad \dfrac{\sin^2 A-1+2\cos A-\cos^2 A}{\sin^2 A+1-2\cos A+\cos^2 A}

\text{Pythagorean Identity:}\qquad \qquad \dfrac{1-\cos^2 A-1+2\cos A}{2-2\cos A}

\text{Simplify:}\qquad \qquad \qquad \dfrac{2\cos A-2\cos^2 A}{2(1-\cos A)}\\\\.\qquad \qquad \qquad \qquad =\dfrac{2\cos A(1-\cos A)}{2(1-\cos A)}

                               =  cos A

LHS = RHS:  cos A = cos A   \checkmark

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