Sag angle Imagine a climber clipping onto the rope described in Example 7 and pulling himself to the rope’s midpoint. Because the rope is supporting the weight of the climber, it no longer takes the shape of the catenary y = 200 cosh x/200. Instead, the rope (nearly) forms two sides of an isosceles triangle. Compute the sag angle θ illustrated in the figure, assuming the rope does not stretch when weighted. Recall from Example 7 that the length of the rope is 101 ft.

Simplify the expression.

$\(\tan\)^2\(\theta\)-\(\sec\)^2\(\theta\)+1$

Simplify the expression.

$\(\frac{\tan\left(-\theta\right)}{\sec\left(-\theta\right)}\)$

Simplify the expression.

$\(\left\)(\(\frac{\tan^2\theta}{\sin^2\theta}\)-1\(\right\))\(\csc\)^2\(\left\)(\(\theta\[\right\))\(\cos\)^2\(\left\)(-\(\theta\]\right\))$

Identify the most helpful first step in verifying the identity.

$\(\left\)(\(\frac{\tan^2\theta}{\sin^2\theta}\)-1\(\right\))=\(\sec\)^2\(\theta\[\sin\)^2\(\left\)(-\(\theta\]\right\))$