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Prove the following trigonometric identity: $\frac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan\frac{\theta}{2}$

TrigonometryTrigonometric IdentitiesTrigonometric EquationsProof
2025/4/29

1. Problem Description

Prove the following trigonometric identity:
1+sin⁡θ−cos⁡θ1+sin⁡θ+cos⁡θ=tan⁡θ2\frac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan\frac{\theta}{2}

2. Solution Steps

We will start with the left-hand side (LHS) of the equation and manipulate it to arrive at the right-hand side (RHS).
We will use the following trigonometric identities:
sin⁡θ=2sin⁡θ2cos⁡θ2\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}
cos⁡θ=cos⁡2θ2−sin⁡2θ2\cos\theta = \cos^2\frac{\theta}{2} - \sin^2\frac{\theta}{2}
1=cos⁡2θ2+sin⁡2θ21 = \cos^2\frac{\theta}{2} + \sin^2\frac{\theta}{2}
Substituting these into the LHS:
LHS=1+sin⁡θ−cos⁡θ1+sin⁡θ+cos⁡θ=(cos⁡2θ2+sin⁡2θ2)+2sin⁡θ2cos⁡θ2−(cos⁡2θ2−sin⁡2θ2)(cos⁡2θ2+sin⁡2θ2)+2sin⁡θ2cos⁡θ2+(cos⁡2θ2−sin⁡2θ2)LHS = \frac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \frac{(\cos^2\frac{\theta}{2} + \sin^2\frac{\theta}{2}) + 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} - (\cos^2\frac{\theta}{2} - \sin^2\frac{\theta}{2})}{(\cos^2\frac{\theta}{2} + \sin^2\frac{\theta}{2}) + 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} + (\cos^2\frac{\theta}{2} - \sin^2\frac{\theta}{2})}
LHS=cos⁡2θ2+sin⁡2θ2+2sin⁡θ2cos⁡θ2−cos⁡2θ2+sin⁡2θ2cos⁡2θ2+sin⁡2θ2+2sin⁡θ2cos⁡θ2+cos⁡2θ2−sin⁡2θ2LHS = \frac{\cos^2\frac{\theta}{2} + \sin^2\frac{\theta}{2} + 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} - \cos^2\frac{\theta}{2} + \sin^2\frac{\theta}{2}}{\cos^2\frac{\theta}{2} + \sin^2\frac{\theta}{2} + 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} + \cos^2\frac{\theta}{2} - \sin^2\frac{\theta}{2}}
LHS=2sin⁡2θ2+2sin⁡θ2cos⁡θ22cos⁡2θ2+2sin⁡θ2cos⁡θ2LHS = \frac{2\sin^2\frac{\theta}{2} + 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2} + 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}
Factor out 2sin⁡θ22\sin\frac{\theta}{2} from the numerator and 2cos⁡θ22\cos\frac{\theta}{2} from the denominator:
LHS=2sin⁡θ2(sin⁡θ2+cos⁡θ2)2cos⁡θ2(cos⁡θ2+sin⁡θ2)LHS = \frac{2\sin\frac{\theta}{2}(\sin\frac{\theta}{2} + \cos\frac{\theta}{2})}{2\cos\frac{\theta}{2}(\cos\frac{\theta}{2} + \sin\frac{\theta}{2})}
LHS=sin⁡θ2cos⁡θ2LHS = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}
LHS=tan⁡θ2LHS = \tan\frac{\theta}{2}
Therefore,
1+sin⁡θ−cos⁡θ1+sin⁡θ+cos⁡θ=tan⁡θ2\frac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan\frac{\theta}{2}

3. Final Answer

tan⁡θ2\tan\frac{\theta}{2}

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