Chapter 7: Q. 101 (page 475)

Show that the functions f and g are identically equal.
$f (θ) = \frac{1 - \sin θ}{\cos θ} - \frac{\cos θ}{1 + \sin θ}, g (θ) = 0$

Short Answer

Expert verified

Yes, the functions are identically equal.

Step by step solution

Step 1. Given Information

The given functions are $f (θ) = \frac{1 - \sin θ}{\cos θ} - \frac{\cos θ}{1 + \sin θ}, g (θ) = 0$

Step 2. Calculation

Simplify any one function and determine whether they are equal or not to the other function.

$f (θ) = \frac{1 - \sin θ}{\cos θ} - \frac{\cos θ}{1 + \sin θ} = \frac{(1 - \sin θ) (1 + \sin θ) - \cos^{2} θ}{\cos θ (1 + \sin θ)} = \frac{1 - \sin^{2} θ - \cos^{2} θ}{\cos θ (1 + \sin θ)} = \frac{1 - (\sin^{2} θ + \cos^{2} θ)}{\cos θ (1 + \sin θ)} = \frac{1 - 1}{\cos θ (1 + \sin θ)} = 0 = g (θ)$

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Show that the functions f and g are identically equal.
$f (θ) = \frac{1 - \sin θ}{\cos θ} - \frac{\cos θ}{1 + \sin θ}, g (θ) = 0$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Calculation

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Show that the functions f and g are identically equal.f(θ)=1-sinθcosθ-cosθ1+sinθ,g(θ)=0

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Calculation

One App. One Place for Learning.

Most popular questions from this chapter

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Study anywhere. Anytime. Across all devices.

Show that the functions f and g are identically equal.
$f (θ) = \frac{1 - \sin θ}{\cos θ} - \frac{\cos θ}{1 + \sin θ}, g (θ) = 0$