Start with the Left Hand Side (LHS) of the identity.

Mathematics

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  1. Start with the Left Hand Side (LHS) of the identity.

    LHS=1+cos⁡θ1−cos⁡θ−1−cos⁡θ1+cos⁡θ\text{LHS} = \frac{1+\cos\theta}{1-\cos\theta} - \frac{1-\cos\theta}{1+\cos\theta}

  2. Find a common denominator and combine the fractions.

    LHS=(1+cos⁡θ)(1+cos⁡θ)−(1−cos⁡θ)(1−cos⁡θ)(1−cos⁡θ)(1+cos⁡θ)\text{LHS} = \frac{(1+\cos\theta)(1+\cos\theta) - (1-\cos\theta)(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)} LHS=(1+cos⁡θ)2−(1−cos⁡θ)21−cos⁡2θ\text{LHS} = \frac{(1+\cos\theta)^2 - (1-\cos\theta)^2}{1-\cos^2\theta}

  3. Expand the numerator and simplify the denominator using the identity 1−cos⁡2θ=sin⁡2θ1-\cos^2\theta = \sin^2\theta.

    LHS=(1+2cos⁡θ+cos⁡2θ)−(1−2cos⁡θ+cos⁡2θ)sin⁡2θ\text{LHS} = \frac{(1+2\cos\theta+\cos^2\theta) - (1-2\cos\theta+\cos^2\theta)}{\sin^2\theta} LHS=1+2cos⁡θ+cos⁡2θ−1+2cos⁡θ−cos⁡2θsin⁡2θ\text{LHS} = \frac{1+2\cos\theta+\cos^2\theta - 1+2\cos\theta-\cos^2\theta}{\sin^2\theta} LHS=4cos⁡θsin⁡2θ\text{LHS} = \frac{4\cos\theta}{\sin^2\theta}

  4. Rewrite the expression to match the Right Hand Side (RHS) using the identity tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}.

    LHS=4cos⁡θsin⁡θ⋅sin⁡θ\text{LHS} = \frac{4\cos\theta}{\sin\theta \cdot \sin\theta} LHS=4sin⁡θ⋅cos⁡θsin⁡θ\text{LHS} = \frac{4}{\sin\theta} \cdot \frac{\cos\theta}{\sin\theta} LHS=4sin⁡θ⋅sin⁡θcos⁡θ\text{LHS} = \frac{4}{\sin\theta \cdot \frac{\sin\theta}{\cos\theta}} LHS=4sin⁡θtan⁡θ\text{LHS} = \frac{4}{\sin\theta \tan\theta}

  5. Conclude that LHS = RHS.

    Since LHS=4sin⁡θtan⁡θ\text{LHS} = \frac{4}{\sin\theta \tan\theta} and RHS=4sin⁡θtan⁡θ\text{RHS} = \frac{4}{\sin\theta \tan\theta}, the identity is proven.

    1+cos⁡θ1−cos⁡θ−1−cos⁡θ1+cos⁡θ≡4sin⁡θtan⁡θ\boxed{\frac{1+\cos\theta}{1-\cos\theta} - \frac{1-\cos\theta}{1+\cos\theta} \equiv \frac{4}{\sin\theta \tan\theta}}
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