Q.Find the values of other five trigonometric functions if $\tan x = -\frac{5}{12}$, $x$ lies in second quadrant.

Concept: Trigonometric Functions in Quadrants — sign of each function depends on the quadrant.

Step 1: In the second quadrant, $\sin x > 0$, $\cos x < 0$, $\tan x < 0$, and $\cot x$, $\sec x$, $\csc x$ follow the signs of their reciprocals.

Step 2: Given $\tan x = -\frac{5}{12} = \frac{\text{opposite}}{\text{adjacent}}$. In the second quadrant, take opposite $= 5$ (positive) and adjacent $= -12$ (negative). Hypotenuse $r = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.

Step 3: Now compute:

• $\sin x = \frac{\text{opposite}}{r} = \frac{5}{13}$
• $\cos x = \frac{\text{adjacent}}{r} = -\frac{12}{13}$ …

Why the quadrant matters first

Trigonometric functions have fixed sign patterns in each quadrant. In the second quadrant ($90^\circ$ to $180^\circ$), only $\sin x$ and its reciprocal $\csc x$ are positive. Everything else — $\cos x$, $\sec x$, $\tan x$, $\cot x$ — is negative. This is not a coincidence: it follows from the unit circle definitions, where $x$ is the angle measured counterclockwise from the positive $x$-axis.

Given $\tan x = -\frac{5}{12}$, the negative sign already tells us we are in either the second or fourth quadrant. The problem explicitly places $x$ in the second quadrant, so we know exactly which signs to assign to each function.

Step-by-step solution

1. Interpret $\tan x$ as a ratio of sides

Recall that $\tan x = \frac{\text{opposite}}{\text{adjacent}} = \frac{y}{x}$ in the coordinate plane. Here, $\tan x = -\frac{5}{12}$ means we can take:

• Opposite side (vertical component) $= 5$ (positive, since $\sin$ is positive in QII)
• Adjacent side (horizontal component) $= -12$ (negative, since $\cos$ is negative in QII)

The hypotenuse $r$ is always positive and found using the Pythagorean theorem:

2. Write $\sin x$ and $\cos x$ from the triangle

From the definitions:

Check: $\tan x = \frac{\sin x}{\cos x} = \frac{5/13}{-12/13} = -\frac{5}{12}$, which matches. Good.

3. Find the reciprocal functions

Reciprocals are straightforward once you have $\sin x$ and $\cos x$:

• $\csc x = \frac{1}{\sin x} = \frac{1}{5/13} = \frac{13}{5}$
• $\sec x = \frac{1}{\cos x} = \frac{1}{-12/13} = -\frac{13}{12}$
• $\cot x = \frac{1}{\tan x} = \frac{1}{-5/12} = -\frac{12}{5}$ …