The problem asks to find a trigonometric ratio of an angle less than 45° that is equal to one of the given trigonometric ratios. The given options are sin 63°, cos 82°, and tan 73°.

1. Problem Description

The problem asks to find a trigonometric ratio of an angle less than 45° that is equal to one of the given trigonometric ratios. The given options are sin 63°, cos 82°, and tan 73°.

2. Solution Steps

We need to find an angle $\theta < 45^\circ$ such that one of the following holds:

sin $\theta$ = sin 63°

cos $\theta$ = cos 82°

tan $\theta$ = tan 73°

Using trigonometric identities:

$\sin x = \sin (180^\circ - x)$. Therefore, $\sin 63^\circ = \sin(180^\circ - 63^\circ) = \sin 117^\circ$. But we are looking for angles less than 45 degrees.

Also, we know that $\cos x = \sin (90^\circ - x)$.

Then, $\cos 82^\circ = \sin (90^\circ - 82^\circ) = \sin 8^\circ$.

Since $8^\circ < 45^\circ$, $\cos 82^\circ$ is a trigonometric ratio of an angle less than $45^\circ$.

For the tangent function, $\tan x = \cot (90^\circ - x) = \frac{1}{\tan (90^\circ - x)}$. Therefore, $\tan x = \frac{1}{\tan(90 - x)}$.

Thus, $\tan 73^\circ = \frac{1}{\tan(90^\circ - 73^\circ)} = \frac{1}{\tan 17^\circ} = \cot 17^\circ$. Also $\tan(x) = \cot(90-x)$ so $\tan 73 = \cot 17$. Since we are comparing to $\tan(\theta)$ values where $\theta<45$, we need to convert to tan values.

$\cot 17 = \tan(90-17) = \tan(73)$ which does not help us since $73 > 45$.

Therefore, $\cos 82^\circ = \sin 8^\circ$. And $8^\circ < 45^\circ$.

3. Final Answer

cos 82° is the trigonometric ratio of an angle less than 45° because cos 82° = sin 8° and 8° < 45°.

So the answer is cos 82°.