Answer:
The approximate value of [tex]f(\theta) = \frac{\sec \theta \cdot (\sec \theta+\tan \theta)}{1+\tan^{2}\theta}[/tex] is 1.366.
Step-by-step explanation:
Let [tex]f(\theta) = \frac{\sec \theta \cdot (\sec \theta+\tan \theta)}{1+\tan^{2}\theta}[/tex], we proceed to simplify the formula until a form based exclusively in sines and cosines is found. From Trigonometry, we shall use the following identities:
[tex]\sec \theta = \frac{1}{\cos \theta}[/tex] (1)
[tex]\tan\theta = \frac{\sin\theta}{\cos \theta}[/tex] (2)
[tex]\cos^{2}+\sin^{2} = 1[/tex] (3)
Then, we simplify the given formula:
[tex]f(\theta) = \frac{\left(\frac{1}{\cos \theta} \right)\cdot \left(\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}\right) }{1+\frac{\sin^{2}\theta}{\cos^{2}\theta} }[/tex]
[tex]f(\theta) = \frac{\left(\frac{1}{\cos^{2} \theta} \right)\cdot (1+\sin \theta)}{\frac{\sin^{2}\theta + \cos^2{\theta}}{\cos^{2}\theta} }[/tex]
[tex]f(\theta) = \frac{\left(\frac{1}{\cos^{2}\theta}\right)\cdot (1+\sin \theta)}{\frac{1}{\cos^{2}\theta} }[/tex]
[tex]f(\theta) = 1+\sin \theta[/tex]
If we know that [tex]\sin \theta =\frac{\sqrt{3}-1}{2}[/tex], then the approximate value of the given function is:
[tex]f(\theta) = 1 +\frac{\sqrt{3}-1}{2}[/tex]
[tex]f(\theta) = \frac{\sqrt{3}+1}{2}[/tex]
[tex]f(\theta) \approx 1.366[/tex]
