Animated Solution for Mathematics - Trigonometry: If $\cot \alpha =1$ and $\sec \beta =-\frac{5}{3}$, where $\pi <\alpha <\frac{3\pi }{2}$ and $\frac{\pi }{2}<\beta <\pi$, then the value of $\tan (\alpha +\beta )$ and the quadrant in which $\alpha +\beta$ lies, respectively are

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a trigonometry problem; we are embarking on a journey through the coordinate plane.

Imagine you are standing at the origin of a standard Cartesian system. The $x$-axis and $y$-axis stretch out before you, dividing the world into four distinct quadrants.

In trigonometry, these quadrants are not mere labels; they are the gatekeepers of reality. They dictate the very essence of our trigonometric ratios—whether they are positive or negative. Before we touch a single equation, we must map our territory.

The Third Quadrant Mystery

Let us look at $\alpha$. We are told that $\pi <\alpha <\frac{3\pi }{2}$.

If you visualize the unit circle, you know that $\pi$ is the negative $x$-axis and $\frac{3\pi }{2}$ is the negative $y$-axis. This means $\alpha$ sweeps through the third quadrant.

In this region, both the $x$ and $y$ coordinates are negative. Consequently, the tangent, which is the ratio of $y$ to $x$, becomes positive.

We are given $\cot \alpha =1$. Since $\tan \alpha$ is the reciprocal of $\cot \alpha$, we find that $\tan \alpha =1$. It is a clean, beautiful start.

The Second Quadrant Challenge

Now, let us turn our attention to $\beta$. The problem places $\beta$ between $\frac{\pi }{2}$ and $\pi$.

This is the second quadrant, a place where the $x$-coordinate is negative and the $y$-coordinate is positive. Here, the tangent must be negative.

We are given $\sec \beta =-\frac{5}{3}$. To find $\tan \beta$, we reach for our trusty identity:

Substituting our value, we get:

Taking the square root, we face a choice: positive or negative $\frac{4}{3}$? Because $\beta$ lives in the second quadrant, we must choose the negative root. Thus, $\tan \beta =-\frac{4}{3}$.

The Heart of the Problem

With $\tan \alpha =1$ and $\tan \beta =-\frac{4}{3}$ in our arsenal, we are ready for the main event. We need to find $\tan (\alpha +\beta )$.

The compound angle formula is our bridge:

Substituting our values, the numerator becomes:

The denominator becomes:

Dividing these, the threes cancel out, leaving us with a final value of $-\frac{1}{7}$.

The Final Quadrant Determination

We have the value, but we need the quadrant. We add the inequalities:

This simplifies to:

This range spans the fourth quadrant and the first quadrant of the next rotation. Since our calculated tangent is $-\frac{1}{7}$, which is negative, the angle must lie in the fourth quadrant, where tangent is negative.

The first quadrant is ruled out because all ratios there are positive. We have arrived at our destination: the value is $-\frac{1}{7}$ and the quadrant is the fourth. You have conquered the problem!