If x, y, z are in A.P. and tan⁻¹ x, tan⁻¹ y and tan⁻¹ z are also in A.P., then (A) x = y = z (B) 2x

Q: Wait, why do we reject the case where $y=0$?

That is a fantastic observation! In our derivation, we canceled $2y$ from both sides. If $y=0$, then $x+z=0$ (from the A.P. condition), meaning $z = -x$. If we plug these into the original condition, we get $\tan^{-1} x, 0, \tan^{-1}(-x)$, which is $\tan^{-1} x, 0, -\tan^{-1} x$. This is indeed an A.P. with a common difference of $-\tan^{-1} x$. So, $y=0$ is a valid solution, but it leads to the trivial case where $x = -z$. However, in the context of the options provided, the general result $x=y=z$ is the intended path.

Q: Why can't we just use the slope of the tangent function to solve this?

You could potentially use calculus, but it would be like using a sledgehammer to crack a nut! The slope of $\tan^{-1} x$ is $\frac{1}{1+x^2}$. While you could analyze the rate of change, the algebraic approach using the identity $\tan^{-1} A + \tan^{-1} B$ is much more direct and less prone to calculation errors. Always look for the algebraic identity first in JEE problems!

Q: How do I know when to use the identity $2\tan^{-1} y = \tan^{-1} \left( \frac{2y}{1-y^2} \right)$?

Whenever you see an expression involving $2\tan^{-1} \theta$ or the sum of two inverse tangents, that is your signal! The inverse tangent addition formula is a staple in JEE trigonometry. Whenever you have a relationship involving inverse trigonometric functions, your first instinct should be to convert them into algebraic equations using these standard identities.

Question

If $x, y, z$ are in A.P. and $	an^{-1} x, 	an^{-1} y$ and $	an^{-1} z$ are also in A.P., then

Options:

(A) x = y = z

(B) 2x = 3y = 6z

(C) 6x = 3y = 2z

(D) 6x = 4y = 3z

Sigma Percentile · Accepted Answer

### The Harmony of Sequences: A Mathematical Journey

My dear student, welcome to a beautiful intersection of algebra and trigonometry. Today, we are not just solving a problem; we are uncovering a hidden symmetry.

We are given three numbers, $x$, $y$, and $z$, that exist in two different worlds simultaneously: the world of Arithmetic Progressions (A.P.) and the world of inverse trigonometric functions. Let us embark on this journey to see how these two worlds collide and what they reveal.

### Phase 1: The Linear Foundation

We begin with the simplest, most elegant structure in mathematics: the Arithmetic Progression. When we say $x, y,$ and $z$ are in A.P., we are saying that the gap between them is constant.

Imagine them as three points on a number line, perfectly spaced. The middle term, $y$, acts as the anchor, the arithmetic mean of its neighbors. Mathematically, this is our bedrock:

$$2y = x + z$$

Let us hold onto this equation. It is the key that will unlock the final door of our problem.

### Phase 2: The Trigonometric Twist

Now, the problem introduces a layer of complexity: $	an^{-1} x, 	an^{-1} y,$ and $	an^{-1} z$ are also in A.P. This means the inverse tangent values themselves are equally spaced.

If we were to plot these on the graph of $f(	heta) = 	an^{-1} 	heta$, we would see that the vertical spacing between the outputs is constant. This gives us our second condition:

$$2	an^{-1} y = 	an^{-1} x + 	an^{-1} z$$

I know what you are thinking—this looks intimidating. But remember, in the JEE, complexity is often just a mask for simplicity. We have a powerful toolkit for inverse trigonometry.

We know that the sum of two inverse tangents can be collapsed into a single term. Specifically, we use the identity:

$$	an^{-1} A + 	an^{-1} B = 	an^{-1} \left( \frac{A+B}{1-AB} ight)$$

And for the left side, we have the double-angle identity for inverse tangents:

$$2	an^{-1} y = 	an^{-1} \left( \frac{2y}{1-y^2} ight)$$

### Phase 3: The Algebraic Collapse

Now, watch the magic happen. By substituting these identities into our second condition, we get:

$$	an^{-1} \left( \frac{2y}{1-y^2} ight) = 	an^{-1} \left( \frac{x+z}{1-xz} ight)$$

Since the inverse tangent function is strictly increasing and one-to-one, we can confidently strip away the $	an^{-1}$ operators. We are left with the equality of the arguments:

$$\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$$

Here is where our earlier work pays off. Recall our bedrock equation: $2y = x + z$. Let us substitute this into the numerator of the right-hand side. The equation becomes:

$$\frac{2y}{1-y^2} = \frac{2y}{1-xz}$$

Do you see it? The numerators are identical! Assuming $y 
eq 0$, we can cancel the $2y$ from both sides. We are left with a simple, elegant relationship:

$$\frac{1}{1-y^2} = \frac{1}{1-xz}$$

This implies $1 - y^2 = 1 - xz$, which simplifies beautifully to $y^2 = xz$.

### Phase 4: The Final Convergence

We have arrived at the destination. We started with $x, y, z$ in A.P., and we have just proven that they must also satisfy $y^2 = xz$, which is the defining condition for a Geometric Progression (G.P.).

Think about the implications. Three numbers that are simultaneously in A.P. and G.P. must be identical. If the common difference is $d$ and the common ratio is $r$, the only way to satisfy both conditions is if $d=0$ and $r=1$.

Thus, **$x = y = z$**.

Mathematics has a way of bringing order to chaos. We started with two seemingly unrelated conditions and found that they force the variables into a state of perfect equality. Keep this logic in your heart, my student. When you face a complex problem, look for the underlying structure, trust your identities, and watch as the complexity collapses into clarity.

Animated Solution for Mathematics - Inverse Trigonometric Functions: If $x, y, z$ are in A.P. and $tan^{- 1} x, tan^{- 1} y$ and $tan^{- 1} z$ are also in A.P., then

Select Answer:

Visualized Solution

The First Arithmetic Progression

The Second Arithmetic Progression

Applying Inverse Trigonometric Identities

Equating the Arguments

Substituting the A.P. Condition

Simplifying the Equation

The Geometric Progression Emerges

The Final Conclusion

The Sigma Insight: Properties of Inverse Trigonometric Functions

The Harmony of Sequences

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Inverse Trigonometric Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x,y,z are in A.P. and tan−1x,tan−1y and tan−1z are also in A.P., then