Question

verify that the equation is an identity
\frac{cos \theta+1}{\tan ^{2} \theta}=\frac{cos \theta}{sec \theta-1}
to verify the identity, start with the more complicated side and transform it to look like the other side choose the correct transformations and transform the expression at each step
\begin{aligned} \frac{cos \theta+1}{\tan ^{2} \theta} &=\\ &=\frac{cos \theta+1}{square} end{aligned}

Explanation:

Step1: Use the identity \(\tan^{2}\theta=\sec^{2}\theta - 1\)

$$ \frac{\cos\theta + 1}{\tan^{2}\theta}=\frac{\cos\theta + 1}{\sec^{2}\theta - 1} $$

Step2: Factor the denominator using \(a^{2}-b^{2}=(a + b)(a - b)\) (here \(a=\sec\theta\), \(b = 1\))

$$ \frac{\cos\theta+1}{\sec^{2}\theta - 1}=\frac{\cos\theta + 1}{(\sec\theta + 1)(\sec\theta - 1)} $$

Step3: Use \(\sec\theta=\frac{1}{\cos\theta}\), then \(\sec\theta + 1=\frac{1+\cos\theta}{\cos\theta}\)

$$ \frac{\cos\theta + 1}{(\sec\theta + 1)(\sec\theta - 1)}=\frac{\cos\theta+1}{\frac{1 + \cos\theta}{\cos\theta}(\sec\theta - 1)} $$

Step4: Simplify the fraction

$$ \frac{\cos\theta+1}{\frac{1 + \cos\theta}{\cos\theta}(\sec\theta - 1)}=\frac{\cos\theta}{\sec\theta - 1} $$

Answer:

The given equation \(\frac{\cos\theta + 1}{\tan^{2}\theta}=\frac{\cos\theta}{\sec\theta - 1}\) is an identity.