Question

Sin∅=√3-1/2 find approximate value of sec∅(sec∅+tan∅)/1+tan²∅​

Answer 1

Answer:

The approximate value of $f(\theta) = \frac{\sec \theta \cdot (\sec \theta+\tan \theta)}{1+\tan^{2}\theta}$ is 1.366.

Step-by-step explanation:

Let $f(\theta) = \frac{\sec \theta \cdot (\sec \theta+\tan \theta)}{1+\tan^{2}\theta}$, 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:

$\sec \theta = \frac{1}{\cos \theta}$ (1)

$\tan\theta = \frac{\sin\theta}{\cos \theta}$ (2)

$\cos^{2}+\sin^{2} = 1$ (3)

Then, we simplify the given formula:

$f(\theta) = \frac{\left(\frac{1}{\cos \theta} \right)\cdot \left(\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}\right) }{1+\frac{\sin^{2}\theta}{\cos^{2}\theta} }$

$f(\theta) = \frac{\left(\frac{1}{\cos^{2} \theta} \right)\cdot (1+\sin \theta)}{\frac{\sin^{2}\theta + \cos^2{\theta}}{\cos^{2}\theta} }$

$f(\theta) = \frac{\left(\frac{1}{\cos^{2}\theta}\right)\cdot (1+\sin \theta)}{\frac{1}{\cos^{2}\theta} }$

$f(\theta) = 1+\sin \theta$

If we know that $\sin \theta =\frac{\sqrt{3}-1}{2}$, then the approximate value of the given function is:

$f(\theta) = 1 +\frac{\sqrt{3}-1}{2}$

$f(\theta) = \frac{\sqrt{3}+1}{2}$

$f(\theta) \approx 1.366$