Answer: see proof below
Explanation:
Use the following Double Angle Identities:
sin 2A = 2cos A · sin A
cos 2A = 2 cos²A - 1
Use the following Quotient Identity: tan A = (sin A)/(cos A)
Use the following Pythagorean Identity:
cos²A + sin²A = 1 --> sin²A = 1 - cos²A
Proof LHS → RHS
Given:
![(1+sin\theta - cos \theta)/(1+sin \theta +cos \theta)](https://img.qammunity.org/2021/formulas/mathematics/high-school/fg6kb2l1ydj1qfivjn43wmhwkgpblnqhke.png)
Let Ф = 2A:
![(1+sin2A - cos 2A)/(1+sin2A +cos2A)](https://img.qammunity.org/2021/formulas/mathematics/high-school/knswef1fi28v0fzl0k48t2o8chjlu6pk28.png)
Un-factor:
![(\bigg((1- cos^2\ 2A)/(1+cos\ 2A)\bigg)+sin\ 2A )/(1+sin\ 2A +cos\ 2A)](https://img.qammunity.org/2021/formulas/mathematics/high-school/owvd8cxlr2icugb1xljcciiuuhructzjas.png)
Pythagorean Identity:
![(\bigg((sin^2\ 2A)/(1+cos\ 2A)\bigg)+sin\ 2A )/(1+cos\ 2A +sin\ 2A)](https://img.qammunity.org/2021/formulas/mathematics/high-school/s1hy290omlrxr878vheuc9of2lu9ffg0eu.png)
Simplify:
![(sin\ 2A)/(1+cos\ 2A)](https://img.qammunity.org/2021/formulas/mathematics/high-school/1sannuvv0ute0gj9z04k0y7bbz77mvblf1.png)
Double Angle Identity:
![(2sin\ A\cdot cos\ A)/(1+(2cos^2 A-1))](https://img.qammunity.org/2021/formulas/mathematics/high-school/cex1fpm0bg7korkiapxzmmsdqnt3p0jf17.png)
Simplify:
![(2sin\ A\cdot cos\ A)/(2cos^2\ A)](https://img.qammunity.org/2021/formulas/mathematics/high-school/71ic2avpfqvfxymj2tsu3v06l856lfelo5.png)
![=(2sin\ A\cdot cos\ A)/(2cos^2\ A)](https://img.qammunity.org/2021/formulas/mathematics/high-school/8rh8sz8x7w9l113618s4pk7hmq55qdkzpt.png)
![=(sin\ A)/(cos\ A)](https://img.qammunity.org/2021/formulas/mathematics/high-school/rg3o7464v3k2k4llfsn0a3ou19h79nwzjl.png)
Quotient Identity: tan A
![\text{Substitute} A = (\theta)/(2)}:\qquad tan(\theta)/(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/dmn4o7ooml63upbtd69x9jf5cq3hvn9b6a.png)
![tan(\theta)/(2) = tan(\theta)/(2)\quad \checkmark](https://img.qammunity.org/2021/formulas/mathematics/high-school/hl9lrcnli1xzk0wlyzn6wr6qjcjv6iij57.png)