JEE Advanced 2004

LEVELJEE Main

Animated Solution for Mathematics - Inverse Trigonometric Functions: The value of $x$ for which $sin (cot^{- 1} (1 + x)) = cos (tan^{- 1} x)$ is

Select Answer:

Visualized Solution

Analyze the Equation

Given equation: $sin (cot^{- 1} (1 + x)) = cos (tan^{- 1} x)$
Goal: Transform inner inverse trigonometric functions to match the outer functions.

Substitution for LHS

Let $θ = cot^{- 1} (1 + x)$
This implies $cot θ = \frac{1 + x}{1}$

Right Triangle for LHS

In a right triangle with angle $θ$ :
$Base = 1 + x$
$Perpendicular = 1$

Hypotenuse of First Triangle

Using Pythagoras theorem:
$Hypotenuse = Base^{2} + Perpendicular^{2}$
$Hypotenuse = (1 + x)^{2} + 1^{2}$

Evaluate $sin (θ)$

We need $sin (θ)$ to match the outer function.
$sin θ = \frac{Perpendicular}{Hypotenuse}$
$sin θ = \frac{1}{1 + ( 1 + x ) ^{2}}$

Substitution for RHS

Let $ϕ = tan^{- 1} x$
This implies $tan ϕ = \frac{x}{1}$

Right Triangle for RHS

In a right triangle with angle $ϕ$ :
$Perpendicular = x$
$Base = 1$

Hypotenuse of Second Triangle

Using Pythagoras theorem:
$Hypotenuse = x^{2} + 1^{2} = 1 + x^{2}$

Evaluate $cos (ϕ)$

We need $cos (ϕ)$ to match the outer function.
$cos ϕ = \frac{Base}{Hypotenuse}$
$cos ϕ = \frac{1}{1 + x ^{2}}$

Equate the Expressions

Original equation: $sin (θ) = cos (ϕ)$
Substitute the derived values:
$\frac{1}{1 + ( 1 + x ) ^{2}} = \frac{1}{1 + x ^{2}}$

Simplify the Equation

Since numerators are equal ( $1 = 1$ ), denominators must be equal:
$1 + (1 + x)^{2} = 1 + x^{2}$

Remove Square Roots

Square both sides to eliminate the square roots:
$1 + (1 + x)^{2} = 1 + x^{2}$

Expand the Binomial

Expand $(1 + x)^{2}$ using $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ :
$1 + (1 + 2 x + x^{2}) = 1 + x^{2}$

Solve for $x$

Simplify the left side: $2 + 2 x + x^{2} = 1 + x^{2}$
Cancel $x^{2}$ from both sides: $2 + 2 x = 1$
Subtract $2$ : $2 x = - 1$
$x = - \frac{1}{2}$

Final Answer

Final Answer: $x = - \frac{1}{2}$
Key Takeaway: Converting inverse trigonometric functions using right-angled triangles simplifies complex equations into basic algebra.

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The Sigma Insight: Solving Inverse Trigonometric Equations

Solution Diagram

Analyzing the Setup

When you see expressions like

sin (cot^{- 1} (1 + x))

, your first instinct might be to panic, but let's pause. Think of the inverse trigonometric function not as a scary operator, but as an angle.

Let

θ = cot^{- 1} (1 + x)

. This is just a fancy way of saying

cot (θ) = 1 + x

Constructing the Triangles

Visualize a right-angled triangle. If the cotangent is the ratio of the base to the perpendicular, then our base is

1 + x

and our perpendicular is

1

By the Pythagorean theorem, the hypotenuse is

(1 + x)^{2} + 1^{2}

. Suddenly, the sine of this angle becomes simple:

sin (θ) = \frac{1}{1 + ( 1 + x ) ^{2}}

We repeat this for the right side. Let

ϕ = tan^{- 1} (x)

, which means

tan (ϕ) = x

Construct another triangle with perpendicular

x

and base

1

. The hypotenuse is

1 + x^{2}

, and thus:

cos (ϕ) = \frac{1}{1 + x ^{2}}

The Master Equation

Equating these two expressions, we get:

\frac{1}{1 + ( 1 + x ) ^{2}} = \frac{1}{1 + x ^{2}}

Squaring both sides yields:

1 + (1 + x)^{2} = 1 + x^{2}

Final Calculation

Expanding the equation, we obtain:

1 + 1 + 2 x + x^{2} = 1 + x^{2}

The

x^{2}

terms vanish, leaving

2 + 2 x = 1

. Solving for

x

2 x = - 1

x = - \frac{1}{2}

The final answer is $x = - 1/2$ .

Common Traps & FAQs

Wait, why do we draw triangles? Can't we just use algebraic identities?

While algebraic identities exist, they often lead to complex domain restrictions and sign errors. Drawing a right-angled triangle is a robust, visual, and foolproof method to convert any inverse trigonometric function into a standard trigonometric ratio. It is the most reliable tool in your JEE arsenal.

Do I need to check the domain of these inverse functions?

In a competitive exam setting, if the equation is given, it is assumed to be defined within its domain. However, for a rigorous understanding, always remember that

cot^{- 1} (u)

and

tan^{- 1} (u)

are defined for all real

u

. Since our final equation involves square roots, we ensure the expressions inside are positive, which they are for all real

x

Why did the

x^{2}

terms cancel out so perfectly?

That is the beauty of well-constructed JEE problems! Often, the algebraic complexity is designed to collapse, rewarding the student who correctly sets up the geometry. If you find terms canceling, it is usually a sign that you are on the right track.

Animated Solution for Mathematics - Inverse Trigonometric Functions: The value of $x$ for which $sin (cot^{- 1} (1 + x)) = cos (tan^{- 1} x)$ is

Select Answer:

Visualized Solution

Analyze the Equation

Substitution for LHS

Right Triangle for LHS

Hypotenuse of First Triangle

Evaluate $sin (θ)$

Substitution for RHS

Right Triangle for RHS

Hypotenuse of Second Triangle

Evaluate $cos (ϕ)$

Equate the Expressions

Simplify the Equation

Remove Square Roots

Expand the Binomial

Solve for $x$

Final Answer

The Sigma Insight: Solving Inverse Trigonometric Equations

Analyzing the Setup

Constructing the Triangles

The Master Equation

Final Calculation

Similar Questions

If $sin^{- 1} (\frac{x}{5}) + cosec^{- 1} (\frac{5}{4}) = \frac{π}{2}$ , then the values of $x$ is

The sum of possible values of $x$ for $tan^{- 1} (x + 1) + cot^{- 1} (\frac{1}{x - 1}) = tan^{- 1} (\frac{8}{31})$ is

$cot^{- 1} (cos α) - tan^{- 1} (cos α) = x$ , then $sin x =$

The number of real solutions of $tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = π /2$ is

Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $tan^{- 1} (x) + tan^{- 1} (2 x) = \frac{π}{4}$ is :

Given that the inverse trigonometric functions take principal values only. Then, the number of real values of $x$ which satisfy $sin^{- 1} (\frac{3 x}{5}) + sin^{- 1} (\frac{4 x}{5}) = sin^{- 1} x$ is equal to:

Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation $cos^{- 1} (x) - 2 sin^{- 1} (x) = cos^{- 1} (2 x)$ is equal to :

If $sin^{- 1} \frac{α}{17} + cos^{- 1} \frac{4}{5} - tan^{- 1} \frac{77}{36} = 0, 0 < α < 13$ , then $sin^{- 1} (sin α) + cos^{- 1} (cos α)$ is equal to

If $k = tan (\frac{π}{4} + \frac{1}{2} cos^{- 1} (\frac{2}{3})) + tan (\frac{1}{2} sin^{- 1} (\frac{2}{3}))$ , then the number of solutions of the equation $sin^{- 1} (k x - 1) = sin^{- 1} x - cos^{- 1} x$ is ......... .

Match List I with List II:

List-I

List-II

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; } The value of x for which sin(cot−1(1+x))=cos(tan−1x) is

Select Answer:

Visualized Solution

Analyze the Equation

Substitution for LHS

Right Triangle for LHS

Hypotenuse of First Triangle

Evaluate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin(θ)

Substitution for RHS

Right Triangle for RHS

Hypotenuse of Second Triangle

Evaluate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos(ϕ)

Equate the Expressions

Simplify the Equation

Remove Square Roots

Expand the Binomial

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Final Answer

The Sigma Insight: Solving Inverse Trigonometric Equations

Analyzing the Setup

Constructing the Triangles

The Master Equation

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If sin−117α​+cos−154​−tan−13677​=0,0<α<13, then sin−1(sinα)+cos−1(cosα) is equal to

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Match List I with List II:

List-I

List-II

Animated Solution for Mathematics - Inverse Trigonometric Functions: The value of $x$ for which $sin (cot^{- 1} (1 + x)) = cos (tan^{- 1} x)$ is

Evaluate $sin (θ)$

Evaluate $cos (ϕ)$

Solve for $x$

If $sin^{- 1} (\frac{x}{5}) + cosec^{- 1} (\frac{5}{4}) = \frac{π}{2}$ , then the values of $x$ is

The sum of possible values of $x$ for $tan^{- 1} (x + 1) + cot^{- 1} (\frac{1}{x - 1}) = tan^{- 1} (\frac{8}{31})$ is

$cot^{- 1} (cos α) - tan^{- 1} (cos α) = x$ , then $sin x =$

The number of real solutions of $tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = π /2$ is

Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $tan^{- 1} (x) + tan^{- 1} (2 x) = \frac{π}{4}$ is :

Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation $cos^{- 1} (x) - 2 sin^{- 1} (x) = cos^{- 1} (2 x)$ is equal to :

If $sin^{- 1} \frac{α}{17} + cos^{- 1} \frac{4}{5} - tan^{- 1} \frac{77}{36} = 0, 0 < α < 13$ , then $sin^{- 1} (sin α) + cos^{- 1} (cos α)$ is equal to

If $k = tan (\frac{π}{4} + \frac{1}{2} cos^{- 1} (\frac{2}{3})) + tan (\frac{1}{2} sin^{- 1} (\frac{2}{3}))$ , then the number of solutions of the equation $sin^{- 1} (k x - 1) = sin^{- 1} x - cos^{- 1} x$ is ......... .