Question

Help me please!!!!!!​

Answer 1

Answer: see both proofs below

Explanation:

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 (A)/(2)=\sqrt{(1+\cos A)/(2)}\\\\\\\sin (A)/(2)=\sqrt{(1-\cos A)/(2)}`

Part 1 Proof LHS → Middle

LHS:
`\tan\bigg((\pi)/(4)-(A)/(2)\bigg)`

Difference Identity:
`(\tan ((\pi)/(4))-\tan((A)/(2)))/(1+\tan((\pi)/(4))\cdot \tan((A)/(2)))`

Unit Circle:
`(1-\tan((A)/(2)))/(1+ \tan((A)/(2)))`

`\text{Tangent Identity:}\qquad \qquad ((\cos(A)/(2)-\sin(A)/(2))/(\cos(A)/(2)))/((\cos(A)/(2)+\sin(A)/(2))/(\cos(A)/(2)))`

Simplify:
`(\cos(A)/(2)-\sin(A)/(2))/(\cos(A)/(2)+\sin(A)/(2))`

LHS = Middle:
`(\cos(A)/(2)-\sin(A)/(2))/(\cos(A)/(2)+\sin(A)/(2))=(\cos(A)/(2)-\sin(A)/(2))/(\cos(A)/(2)+\sin(A)/(2))\qquad \checkmark`

Part 2 Proof Middle → RHS

Middle:
`(\cos(A)/(2)-\sin(A)/(2))/(\cos(A)/(2)+\sin(A)/(2))`

`\text{Half-Angle Identity:}\qquad \qquad \frac{\sqrt{(1+\cos A)/(2)}-\sqrt{(1-\cos A)/(2)}}{\sqrt{(1+\cos A)/(2)}+\sqrt{(1-\cos A)/(2)}}`

Simplify:
`(√(1+\cos A)-√(1-\cos A))/(√(1+\cos A)+√(1-\cos A))`

Rationalize Denominator:
`(√(1+\cos A)-√(1-\cos A))/(√(1+\cos A)+√(1-\cos A))\bigg((√(1+\cos A)-√(1-\cos A))/(√(1+\cos A)-√(1-\cos A))\bigg)`

`=(1+\cos A-2√(1-\cos^2 A)+1-\cos A)/(1+\cos A-(1-\cos A))`

Simplify:
`(2-2√(1-\cos^2 A))/(2\cos A)`

`=(2-2√(sin^2 A))/(2\cos A)`

`= (2-2\sin A)/(2\cos A)`

Factor:
`(2(1-\sin A))/(2(\cos A))`

Simplify:
`(1-\sin A)/(\cos A)`

Expand:
`(1-\sin A)/(\cos A)\bigg((1+\sin A)/(1+\sin A)\bigg)`

`=(1-\sin^2 A)/(\cos A(1+\sin A))`

Simplify:
`(\cos^2 A)/(\cos A(1+\sin A))`

`=(\cos A)/(1+\sin A)`

Middle = RHS:
`(\cos A)/(1+\sin A)=(\cos A)/(1+\sin A)\qquad \checkmark`