Q.Find the values of other five trigonometric functions if $\cot x = \frac{4}{3}$, $x$ lies in third quadrant.

Concept: Trigonometric Functions in Quadrants — in the third quadrant, only $\tan$ and $\cot$ are positive; $\sin$, $\cos$, $\sec$, $\csc$ are negative.

Step 1: Given $\cot x = \frac{4}{3}$, so $\tan x = \frac{1}{\cot x} = \frac{3}{4}$.

Step 2: Represent $\cot x = \frac{\text{adjacent}}{\text{opposite}} = \frac{4}{3}$. In the third quadrant, both adjacent and opposite are negative, so take $OA = -4$, $OP = -3$. Hypotenuse $OM = \sqrt{(-4)^2 + (-3)^2} = 5$.

Step 3: Now compute:

• $\sin x = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{-3}{5}$
• $\cos x = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{-4}{5}$
• $\csc x = \frac{1}{\sin x} = -\frac{5}{3}$
• $\sec x = \frac{1}{\cos x} = -\frac{5}{4}$

Concept and Intuition

When solving for trigonometric functions given one function and the quadrant, the key is understanding sign conventions in each quadrant. In the third quadrant ($\pi < x < \frac{3\pi}{2}$), both sine and cosine are negative. Since $\cot x = \frac{\cos x}{\sin x}$, a positive $\cot x$ here means both $\cos x$ and $\sin x$ are negative (negative divided by negative gives positive). This sign awareness prevents the most common mistake: forgetting that the Pythagorean theorem gives only the magnitude, not the sign.

Step-by-Step Solution

1. Find $\tan x$ from $\cot x$

Since $\tan x$ and $\cot x$ are reciprocals:

2. Interpret $\cot x$ as a ratio of sides

$\cot x = \frac{4}{3}$ means that in a right triangle (ignoring signs for a moment), the adjacent side to angle $x$ is $4$ and the opposite side is $3$. But remember: this is just the magnitude — the actual signs depend on the quadrant.

3. Find the hypotenuse using the Pythagorean theorem

So the three sides of the reference triangle are: adjacent $= 4$, opposite $= 3$, hypotenuse $= 5$.

4. Determine $\cos x$ and $\sin x$ with correct signs

In the third quadrant, both $\cos x$ and $\sin x$ are negative.

• $\cos x = -\frac{\text{adjacent}}{\text{hypotenuse}} = -\frac{4}{5}$
• $\sin x = -\frac{\text{opposite}}{\text{hypotenuse}} = -\frac{3}{5}$

5. Find $\sec x$ and $\csc x$ as reciprocals

• $\sec x = \frac{1}{\cos x} = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}$
• $\csc x = \frac{1}{\sin x} = \frac{1}{-\frac{3}{5}} = -\frac{5}{3}$

6. Verify consistency

Check that $\cot x = \frac{\cos x}{\sin x} = \frac{-4/5}{-3/5} = \frac{4}{3}$, which matches the given value. This confirms our signs are correct.