Understanding the Trigonometry Problem

The question asks us to find the value of tan x given that cos x = p/q and that the angle x is between 0° and 90°. The condition 0° < x < 90° means that the angle x lies in the first quadrant. In the first quadrant, all basic trigonometric ratios (sine, cosine, tangent, cotangent, secant, cosecant) are positive.

Relating cos x to a Right Triangle

We know that for an acute angle x in a right-angled triangle, the cosine of the angle is defined as the ratio of the length of the adjacent side to the length of the hypotenuse.

So, given $\cos x = \frac{p}{q}$, we can imagine a right-angled triangle where:

• The side adjacent to angle x has a length proportional to p.
• The hypotenuse has a length proportional to q.

Let's assume the adjacent side is exactly p and the hypotenuse is exactly q for simplicity in calculation. Since x is in the first quadrant, cos x must be positive, which implies that p and q must have the same sign (we typically consider side lengths as positive, so p > 0 and q > 0).

Using the Pythagorean Theorem

To find tan x, which is defined as the ratio of the opposite side to the adjacent side, we first need to find the length of the side opposite to angle x. We can use the Pythagorean theorem in the right-angled triangle.

The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides (adjacent and opposite).

Let the opposite side be denoted by o.

$\text{(Adjacent)}^2 + \text{(Opposite)}^2 = \text{(Hypotenuse)}^2$

$p^2 + o^2 = q^2$

Now, we solve for $o^2$:

$o^2 = q^2 - p^2$

To find the length of the opposite side o, we take the square root of both sides:

$o = \sqrt{q^2 - p^2}$

Since x is in the first quadrant, the opposite side length must be positive. For the square root to be a real number, we must have $q^2 - p^2 \ge 0$, which means $q^2 \ge p^2$. Since q is the hypotenuse and p is a side, it is always true that $q \ge p$ in a right triangle.

Calculating tan x

Now that we have the lengths of the opposite side and the adjacent side, we can calculate tan x.

$\tan x = \frac{\text{Opposite}}{\text{Adjacent}}$

Substitute the values we found:

$\tan x = \frac{\sqrt{q^2 - p^2}}{p}$

Summary of Steps

Here’s a quick recap of the steps followed to find the value of tan x:

1. Identify the given information: $\cos x = \frac{p}{q}$ and $0° < x < 90°$.
2. Relate $\cos x$ to a right-angled triangle (Adjacent = p, Hypotenuse = q).
3. Use the Pythagorean theorem ($p^2 + o^2 = q^2$) to find the opposite side $o = \sqrt{q^2 - p^2}$.
4. Calculate $\tan x$ using the ratio $\frac{\text{Opposite}}{\text{Adjacent}}$.

The calculated value of $\tan x$ is $\frac{\sqrt{q^2 - p^2}}{p}$.

Revision Table: Key Trigonometric Concepts

Additional Information on Trigonometric Identities

Besides using a right triangle, you can also use trigonometric identities to solve such problems. One fundamental identity is:

$\sin^2 x + \cos^2 x = 1$

Given $\cos x = \frac{p}{q}$, we can find $\sin x$:

$\sin^2 x + \left(\frac{p}{q}\right)^2 = 1$

$\sin^2 x = 1 - \frac{p^2}{q^2} = \frac{q^2 - p^2}{q^2}$

Since $x$ is in the first quadrant ($\sin x > 0$), we take the positive square root:

$\sin x = \sqrt{\frac{q^2 - p^2}{q^2}} = \frac{\sqrt{q^2 - p^2}}{\sqrt{q^2}} = \frac{\sqrt{q^2 - p^2}}{|q|}$

Since $0° < x < 90°$, $q$ is typically considered positive (as it represents a hypotenuse length), so $|q| = q$.

$\sin x = \frac{\sqrt{q^2 - p^2}}{q}$

Now, using the identity $\tan x = \frac{\sin x}{\cos x}$:

$\tan x = \frac{\frac{\sqrt{q^2 - p^2}}{q}}{\frac{p}{q}} = \frac{\sqrt{q^2 - p^2}}{q} \times \frac{q}{p} = \frac{\sqrt{q^2 - p^2}}{p}$

Both methods yield the same result, confirming the value of $\tan x$ is $\frac{\sqrt{q^2 - p^2}}{p}$.