Sigma Percentile
JEE Main 2021 (20 July Shift 2)
LEVELJEE Main

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

Select Answer:

Visualized Solution

Objective

  • Evaluate
  • Convert all terms to

Formula for

  • Identity:
  • Here,

Substituting

  • Substitute :

Simplifying the Expression

  • Numerator:
  • Denominator:
  • Result:

Setting up the Triangle

  • Let
  • In a right triangle: Perpendicular , Hypotenuse

Calculating the Base

  • Using Pythagoras Theorem:

Converting to

  • Therefore,

Updated Expression

  • The expression becomes:

Using Identity

  • Formula:
  • Let and

Substituting Values

  • Substitute values into the formula:

Solving the Numerator

  • Numerator calculation:

Solving the Denominator

  • Denominator calculation:

Final Calculation

  • Final Division:

The Sigma Insight: Properties of Inverse Trigonometric Functions

Solution Diagram

Analyzing the Setup

Imagine you are standing at the threshold of a complex trigonometric landscape. You are presented with the expression .
At first glance, it looks like a chaotic mix of inverse functions. However, our goal is to transform this expression into a language that the outer function understands perfectly.
We want everything inside that bracket to be a function. Because , this is the ultimate simplification.

Taming the First Term

Let us focus on the first part: . We have a coefficient of that is blocking our path.
We reach into our toolkit and pull out the double-angle identity for inverse tangent:
By substituting , the expression becomes:
The numerator is , and the denominator is . When we divide by , we are effectively multiplying by .
Simplifying this, we are left with . The first beast is tamed.

The Geometry of the Triangle

Now, we turn to . This is where we use our geometric intuition.
Let , which implies . Picture a right-angled triangle where the perpendicular is and the hypotenuse is .
Using the Pythagorean theorem, the base is:
Since , we find that . Thus, . We have achieved total uniformity.

The Grand Finale

Our expression is now . We apply the tangent addition formula:
Here, and . Substituting these, we get:
Calculating the numerator:
Calculating the denominator:
Finally, we divide the results:
We have arrived at the summit. The complexity has dissolved into the final answer: .

Similar Questions

JEE Advanced 1983
LEVELJEE Main

The value of is

(A)
6/17
(B)
7/16
(C)
16/7
(D)
none
JEE Main 2026 (28 January Shift 2)
LEVELJEE Main

Considering the principal values of inverse trigonometric functions, the value of the expression is equal to :

(A)
(B)
(C)
(D)
JEE Main 2022 (26 July Shift 1)
LEVELJEE Main

is equal to:

(A)
1
(B)
2
(C)
(D)
JEE Advanced 1994
LEVELJEE Main

If we consider only the principle values of the inverse trigonometric functions then the value of is

(A)
(B)
(C)
(D)
JEE(ADVANCED)-202
LEVELJEE Main

Considering only the principal values of the inverse trigonometric functions, the value of is

(A)
(B)
(C)
(D)
JEE Main 2013
LEVELJEE Main

The value of is

(A)
6/17
(B)
3/17
(C)
4/17
(D)
5/17
JEE Main 2019 (12 April Shift 1)
LEVELJEE Main

The value of is equal to :

(A)
(B)
(C)
(D)
JEE Advanced 1984
LEVELBoard

The numerical value of is equal to .........

JEE Main 2023 (24 January Shift 1)
LEVELBoard

is equal to

(A)
(B)
(C)
(D)
JEE Main 2025 (January)
LEVELJEE Main

is equal to:

(A)
1
(B)
0
(C)
(D)