Question

Given sec A= and that angle A is in Quadrant I, find the exact value of csc A in
simplest radical form using a rational denominator.
Need help :(

Answer 1

The required value is $\csc A=\dfrac{5}{4}$.

Important information:

• $\sec A=\dfrac{5}{3}$
• Angle A is in Quadrant I.

We need to find the exact value of csc A.

Trigonometric Ratios:

Formulae used: If angle A is in Quadrant I.

$\cos A=\dfrac{1}{\sec A}$

$\sin A=\sqrt{1-\cos^2 A}$

$\csc A=\dfrac{1}{\sin A}$

Using these formulae, we get

$\cos A=\dfrac{3}{5}$

And,

$\sin A=\sqrt{1-\left(\dfrac{3}{5}\right)^2}$

$\sin A=\sqrt{1-\dfrac{9}{25}}$

$\sin A=\sqrt{\dfrac{16}{25}}$

$\sin A=\dfrac{4}{5}$

Now,

$\csc A=\dfrac{1}{\sin A}$

$\csc A=\dfrac{1}{\frac{4}{5}}$

$\csc A=\dfrac{5}{4}$

Thus, the required value is $\csc A=\dfrac{5}{4}$.

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Answer 2

The required value of cscA is 1.25

Given,

$secA=\frac{5}{3}$

Trigonometrical ratios:

In the first quadrant, all ratios are positives.

$sin A=\frac{P}{H}$

And the value of hypotenuse is,

$P=\sqrt{H^2-B^2} \\=\sqrt{5^2-3^2} \\=\sqrt{16}\\ P=4$

So, the sine A is,

$sinA=\frac{P}{H}\\ =\frac{4}{5}$

So, the rational denominator is,

$cscA=\frac{H}{P}\\ =\frac{5}{4}\\=1.25$

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