Solving Trigonometry Problems: Finding tan x / cot x

We are given the value of $\sin x = \frac{4}{5}$ and asked to find the value of $\frac{{\tan x}}{{\cot x}}$. To solve this trigonometry problem, we first need to find the values of $\tan x$ and $\cot x$ using the given information.

Using Trigonometric Identities

We know the fundamental trigonometric identity:

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

We can use this identity to find the value of $\cos x$. Substitute the given value of $\sin x$:

$\left(\frac{4}{5}\right)^2 + \cos^2 x = 1$

$\frac{16}{25} + \cos^2 x = 1$

Now, solve for $\cos^2 x$:

$\cos^2 x = 1 - \frac{16}{25}$

To subtract the fractions, find a common denominator:

$\cos^2 x = \frac{25}{25} - \frac{16}{25}$

$\cos^2 x = \frac{25 - 16}{25}$

$\cos^2 x = \frac{9}{25}$

Taking the square root of both sides gives $\cos x = \pm \sqrt{\frac{9}{25}} = \pm \frac{3}{5}$.

Since the options for $\frac{\tan x}{\cot x}$ are positive, we assume $x$ is in a quadrant where $\tan x$ and $\cot x$ have the same sign (either both positive or both negative). If we assume $x$ is in the first quadrant, then $\sin x$ and $\cos x$ are both positive. So, we take $\cos x = \frac{3}{5}$.

Calculating tan x and cot x

Now that we have $\sin x$ and $\cos x$, we can find $\tan x$ using the identity $\tan x = \frac{\sin x}{\cos x}$.

$\tan x = \frac{\frac{4}{5}}{\frac{3}{5}}$

$\tan x = \frac{4}{5} \times \frac{5}{3}$

$\tan x = \frac{4}{3}$

Next, we find $\cot x$ using the identity $\cot x = \frac{1}{\tan x}$.

$\cot x = \frac{1}{\frac{4}{3}}$

$\cot x = \frac{3}{4}$

Finding the Ratio tan x / cot x

Finally, we calculate the required ratio $\frac{\tan x}{\cot x}$:

$\frac{\tan x}{\cot x} = \frac{\frac{4}{3}}{\frac{3}{4}}$

Dividing by a fraction is the same as multiplying by its reciprocal:

$\frac{\tan x}{\cot x} = \frac{4}{3} \times \frac{4}{3}$

$\frac{\tan x}{\cot x} = \frac{16}{9}$

Alternatively, we know that $\cot x = \frac{1}{\tan x}$. So, $\frac{\tan x}{\cot x} = \frac{\tan x}{\frac{1}{\tan x}} = \tan x \times \tan x = \tan^2 x$.

Since we found $\tan x = \frac{4}{3}$, we can calculate $\tan^2 x$ directly:

$\tan^2 x = \left(\frac{4}{3}\right)^2 = \frac{4^2}{3^2} = \frac{16}{9}$

Both methods give the same result.

The value of $\frac{{\tan x}}{{\cot x}}$ is $\frac{16}{9}$.

Revision Table: Key Trigonometry Concepts

Additional Information: Trigonometric Ratios and Quadrants

When solving trigonometry problems, the quadrant in which the angle lies is important because it determines the sign of the trigonometric ratios. For example:

• Quadrant I (0° to 90°): All trigonometric ratios (sin, cos, tan, cot, sec, csc) are positive.
• Quadrant II (90° to 180°): Sine and cosecant are positive; others are negative.
• Quadrant III (180° to 270°): Tangent and cotangent are positive; others are negative.
• Quadrant IV (270° to 360°): Cosine and secant are positive; others are negative.

In this specific problem, given $\sin x = \frac{4}{5}$ (positive), $x$ could be in Quadrant I or II. However, the final answer options are positive, suggesting that $\tan x$ and $\cot x$ must have the same sign. This happens in Quadrant I (both positive) and Quadrant III (both negative). If $x$ were in Quadrant III, $\cos x$ would be negative ($\frac{-3}{5}$), $\tan x$ would be $\frac{4/5}{-3/5} = -\frac{4}{3}$, and $\cot x$ would be $\frac{1}{-4/3} = -\frac{3}{4}$. The ratio $\frac{\tan x}{\cot x}$ would then be $\frac{-4/3}{-3/4} = \frac{16}{9}$, which is the same positive result. Therefore, the calculation $\frac{\tan x}{\cot x} = \tan^2 x$ confirms that the result is always positive, regardless of the quadrant (as long as $\tan x$ is defined). Our assumption of the first quadrant yields the correct numerical value for the ratio.