Question

Please help me!!!!!​

Answer 1

Answer: see proof below

Explanation:

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

Proof LHS → RHS

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

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

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

`\text{Simplify:}\qquad \qquad \qquad (1-\tan^2(A)/(2))/(1+\tan^2(A)/(2))`

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

`\text{Simplify:}\qquad \qquad (\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 (1-\cos^2 A-1+2\cos A)/(2-2\cos A)`

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

= cos A

LHS = RHS: cos A = cos A
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