(5) \( \frac{1}{1-\cos \theta}+\frac{1}{1+\cos \theta}=\frac{2}{\sin ^{2} \theta} \)
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Let’s start transforming the left-hand side of the equation: \[ \frac{1}{1-\cos \theta} + \frac{1}{1+\cos \theta} \] To combine these fractions, we find a common denominator, which is \((1 - \cos \theta)(1 + \cos \theta) = 1 - \cos^2 \theta = \sin^2 \theta\) (using the Pythagorean identity). Thus: \[ \frac{1 \cdot (1+\cos\theta) + 1 \cdot (1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)} = \frac{(1+\cos\theta) + (1-\cos\theta)}{\sin^2 \theta} = \frac{2}{\sin^2 \theta} \] Now we equate the left-hand side to the right-hand side: \[ \frac{2}{\sin^2 \theta} = \frac{2}{\sin^2 \theta} \] This equality is satisfied for all \(\theta\) where \(\sin^2 \theta \neq 0\) (i.e., \(\theta \neq k\pi\) for \(k \in \mathbb{Z}\)). Hence, we conclude that the given equation is true for valid \(\theta\). Therefore, the relation holds, confirming the equation: \[ \frac{1}{1-\cos \theta}+\frac{1}{1+\cos \theta}=\frac{2}{\sin ^{2} \theta} \] is valid under the condition stated.
