Question

establish the identity
\\(\frac{\sec\theta - \tan\theta}{\sec\theta\tan\theta}=\cot\theta - \cos\theta\\)

write the left side as a difference of two quotients
\\(\frac{\square}{\sec\theta\tan\theta}-\frac{\square}{\sec\theta\tan\theta}\\)

Explanation:

Step1: Split the left - hand side fraction

We know that $\frac{a - b}{c}=\frac{a}{c}-\frac{b}{c}$. So, $\frac{\sec\theta-\tan\theta}{\sec\theta\tan\theta}=\frac{\sec\theta}{\sec\theta\tan\theta}-\frac{\tan\theta}{\sec\theta\tan\theta}$.

Step2: Simplify each quotient

For the first quotient $\frac{\sec\theta}{\sec\theta\tan\theta}$, since $\sec\theta
eq0$ and $\tan\theta
eq0$, we can cancel out $\sec\theta$ and get $\frac{1}{\tan\theta}=\cot\theta$. For the second quotient $\frac{\tan\theta}{\sec\theta\tan\theta}$, we can cancel out $\tan\theta$ and get $\frac{1}{\sec\theta}=\cos\theta$.

Step3: Combine the simplified quotients

$\frac{1}{\tan\theta}-\frac{1}{\sec\theta}=\cot\theta - \cos\theta$, which is the right - hand side of the given identity.

Answer:

The identity $\frac{\sec\theta-\tan\theta}{\sec\theta\tan\theta}=\cot\theta - \cos\theta$ is established.