Question

Given csc(A) = 60/16 and that angle A is in Quadrant I, find the exact value of sec A in simplest radical form using a rational denominator . Someone please help me!

Answer 1

Answer:

$\frac{15 \sqrt{209} }{209}$

Step-by-step explanation:

Objective: Understand and work with trig identies.

Recall multiple trig identies and manipulate them to get from cosecant to secant.

Given

$\csc(a) = \frac{60}{16}$

Apply reciprocal identity csc a = 1/sin a.

$\sin(a) = \frac{16}{60}$

Apply pythagorean identity to find cos a.

$( \frac{16}{60}) {}^{2} + \cos(x) {}^{2} = 1$

$\frac{256}{3600} + \cos(x) {}^{2} = 1$

$\cos(x) {}^{2} = \frac{3600}{3600} - \frac{256}{3600}$

We can simplify both expression

$\cos(x) {}^{2} = \frac{225}{225} - \frac{16}{225}$

$\cos(x) = \frac{ \sqrt{209} }{15}$

Cosine is positve on quadrant 1 so that cos(a)

Apply reciprocal identity sec a= 1/ cos a.

The answer is

$\sec(a) = \frac{15}{ \sqrt{209} }$

Rationalize the denominator.

$\frac{15}{ \sqrt{209} } \times \frac{ \sqrt{209} }{ \sqrt{209} } = \frac{15 \sqrt{209} }{209}$