Chapter 7: Q. 102 (page 475)

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

Short Answer

Expert verified

Yes, the functions are identically equal.

Step by step solution

Step 1. Given Information

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

Step 2. Calculation

Simplify both the functions and determine whether they are equal or not to each other.

$f (θ) = \tan θ + s e c θ = \frac{\sin θ}{\cos θ} + \frac{1}{\cos θ} = \frac{1 + \sin θ}{\cos θ}$

And

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

Thus, both the functions are equal.

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

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Calculation

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