Sigma Percentile
JEE Advanced 1984
LEVELBoard

Animated Solution for Mathematics - Inverse Trigonometric Functions: The numerical value of is equal to .........

Visualized Solution

Analyze the Expression

  • Given expression:
  • Let's define the inner angle:
  • This implies:

Visualizing the Angle

  • We can represent geometrically on a coordinate plane.
  • Since , we draw a line from the origin to the point .
  • The angle formed with the positive x-axis is .

The Double Angle Identity

  • To simplify , we use the double-angle identity.
  • Identity:

Substituting into the Identity

  • Substitute into the identity:

Simplifying Numerator and Denominator

  • Numerator:
  • Denominator:

Calculating

  • Divide the numerator by the denominator:
  • Therefore,

Comparing with

  • Let's plot and on the coordinate plane.
  • Since and , we have .
  • This means the final angle will be negative.

The Difference Formula for Tangent

  • We need to evaluate .
  • Using the identity:

Substituting the Values

  • Substitute and :
  • Substitute and :

Simplifying the Fraction

  • Numerator:
  • Denominator:

Arriving at the Final Value

  • Divide the numerator by the denominator:
  • The final numerical value is .

The Sigma Insight: Properties of Inverse Trigonometric Functions

Solution Diagram
Welcome, future engineer. Today, we are going to dismantle a problem that often intimidates students, not because it is inherently difficult, but because it tests your ability to remain calm and systematic in the face of composite functions.
We are looking at the expression . When you see this, your first instinct might be to panic about the inverse tangent. But I want you to take a deep breath. Mathematics is not about memorizing formulas; it is about recognizing patterns.

Phase 1

The Deconstruction
Let us define our territory. We have an inner angle, . By definition, this means .
Imagine this on a coordinate plane. You are standing at the origin, looking at a point . The angle your line of sight makes with the positive -axis is . It is a small, sharp angle.
Now, the expression asks us to evaluate the tangent of . We have successfully reduced the complexity of the problem. We are no longer dealing with inverse functions; we are dealing with a simple subtraction of angles.

Phase 2

The Bridge
Now, we face the term . How do we find the tangent of a doubled angle when we only know the tangent of the single angle?
This is where the double-angle identity for inverse tangent becomes our best friend. The identity states:
Think of this as a bridge. It allows us to cross from the world of inverse functions into the world of standard algebraic fractions. By substituting , we get:
Let us pause here. Do not rush the arithmetic. The numerator is . The denominator is , which is .
When we divide these, we get:
So, we have discovered that .

Phase 3

The Geometric Reality
Before we proceed to the final step, let us perform a sanity check. We know , which is approximately . We also know that .
Since , it is geometrically certain that . This confirms that our final result must be negative.
If you ever get a positive answer, you know immediately that you have made a sign error. This is the power of geometric intuition—it acts as a safety net for your algebra.

Phase 4

The Final Calculation
We are now at the finish line. We need to evaluate . We invoke the subtraction formula:
Substituting and , we get:
Look at the elegance of this expression. The numerator becomes . The denominator becomes .
When we divide these, the common denominator of cancels out beautifully, leaving us with .
You see? The complexity melts away when you follow the logic. You have just navigated a multi-step trigonometric problem with precision. Keep this confidence. Every time you face a complex expression, break it down, find the bridge, and trust the process. You are ready for the next challenge.

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