$\sqrt{1+x^2} \left[ \{x \cos (\cot^{-1} x) + \sin (\cot^{-1} x)\}^2 - 1 \right]^{1/2} =$ Options: (A) $\frac{x}{\sqrt{1+x^2}}$ (B) x (C) x$\sqrt{1+x^2}$ (D) $\sqrt{1+x^2}$

### Analyzing the Setup The expression $\sqrt{1+x^2} \left[ \{x \cos (\cot^{-1} x) + \sin (\cot^{-1} x)\}^2 - 1 \right]^{1/2}$ appears complex, but we can simplify it using trigonometric substitution. Let us define the angle $\theta = \cot^{-1} x$. This implies that $\cot \theta = x$, or more helpfully, $\cot \theta = \frac{x}{1}$. ### Building the Foundation Consider a right-angled triangle where $\theta$ is an acute angle. By the definition of the cotangent function, we set the base to $x$ and the perpendicular to $1$. Using the Pythagorean theorem, the hypotenuse is calculated as: $$H = \sqrt{x^2 + 1^2} = \sqrt{1+x^2}$$ From this triangle, we derive the following trigonometric ratios: $$\cos \theta = \frac{x}{\sqrt{1+x^2}}, \quad \sin \theta = \frac{1}{\sqrt{1+x^2}}$$ ### Taming the Inner Beast Now, focus on the inner bracket of the original expression: $x \cos \theta + \sin \theta$. Substituting our geometric values, we get: $$x \left( \frac{x}{\sqrt{1+x^2}} \right) + \left( \frac{1}{\sqrt{1+x^2}} \right) = \frac{x^2 + 1}{\sqrt{1+x^2}}$$ Since $x^2 + 1 = (\sqrt{1+x^2})^2$, the expression simplifies significantly: $$\frac{(\sqrt{1+x^2})^2}{\sqrt{1+x^2}} = \sqrt{1+x^2}$$ ### The Final Unraveling We now substitute this result back into the original expression. The term inside the square brackets becomes: $$\left[ (\sqrt{1+x^2})^2 - 1 \right]^{1/2} = \left[ 1 + x^2 - 1 \right]^{1/2} = \sqrt{x^2} = |x|$$ Assuming $x > 0$, the expression simplifies to $\sqrt{1+x^2} \cdot x$. The final simplified result is **$x\sqrt{1+x^2}$**.

Wait, why do we assume $\cot \theta = x/1$? Is there a reason for the denominator?

That is a fantastic question! In trigonometry, we often work with ratios. By writing $\cot \theta = x$ as $\frac{x}{1}$, we are simply making the ratio explicit: $\frac{\text{Base}}{\text{Perpendicular}}$. This makes it trivial to identify the base as $x$ and the perpendicular as $1$, allowing us to construct our triangle and find the hypotenuse $\sqrt{x^2+1}$ immediately.

Why does the inner expression simplify so cleanly to $\sqrt{1+x^2}$?

It is not magic, it is the beauty of mathematics! When you substitute $\cos \theta$ and $\sin \theta$ into $x \cos \theta + \sin \theta$, you get $\frac{x^2}{\sqrt{1+x^2}} + \frac{1}{\sqrt{1+x^2}}$. Combining these gives $\frac{x^2+1}{\sqrt{1+x^2}}$. Since $x^2+1$ is just $(\sqrt{1+x^2})^2$, dividing it by $\sqrt{1+x^2}$ leaves you with exactly $\sqrt{1+x^2}$. It is a perfect cancellation designed to reward your persistence.

JEE Advanced 2008

LEVELJEE Main

Animated Solution for Mathematics - Inverse Trigonometric Functions: $1 + x^{2} [{x cos (cot^{- 1} x) + sin (cot^{- 1} x)}^{2} - 1]^{1/2} =$

Q: Can I just use trigonometric identities instead of drawing a triangle?

You absolutely can! You could use identities like $\cos(\cot^{-1} x) = \frac{x}{\sqrt{1+x^2}}$ and $\sin(\cot^{-1} x) = \frac{1}{\sqrt{1+x^2}}$. However, drawing the triangle is a powerful, foolproof method that prevents memory errors and helps you visualize the relationship between the sides. It is a tool that never fails, even when you forget the specific identity.

Select Answer:

Visualized Solution

Defining $θ$

Let $θ = cot^{- 1} x$
This implies $cot θ = x$
We can write this as $cot θ = \frac{x}{1}$

Visualizing the Right Triangle

In a right-angled triangle, $cot θ = \frac{Base}{Perpendicular}$
Base $= x$
Perpendicular $= 1$

Calculating the Hypotenuse

Using Pythagoras Theorem:
$Hypotenuse = Base^{2} + Perpendicular^{2}$
$Hypotenuse = x^{2} + 1^{2} = 1 + x^{2}$

Finding $cos θ$

$cos θ = \frac{Base}{Hypotenuse}$
$cos θ = \frac{x}{1 + x ^{2}}$

Finding $sin θ$

$sin θ = \frac{Perpendicular}{Hypotenuse}$
$sin θ = \frac{1}{1 + x ^{2}}$

Substituting into the Inner Expression

Inner expression: $x cos θ + sin θ$
Substitute the values:
$= x (\frac{x}{1 + x ^{2}}) + (\frac{1}{1 + x ^{2}})$

Simplifying the Numerator

Multiply the terms:
$= \frac{x ^{2}}{1 + x ^{2}} + \frac{1}{1 + x ^{2}}$
Combine the fractions:
$= \frac{x ^{2} + 1}{1 + x ^{2}}$

Simplifying the Fraction

Notice that $x^{2} + 1$ is the square of $x^{2} + 1$
So, $\frac{x ^{2} + 1}{1 + x ^{2}} = 1 + x^{2}$

Squaring the Result

The expression has a square on the inner term:
$(1 + x^{2})^{2}$
$= 1 + x^{2}$

Subtracting $1$

Now, subtract $1$ as per the expression:
$(1 + x^{2}) - 1$
$= x^{2}$

Applying the Outer Power

Apply the power of $\frac{1}{2}$ (which is the square root):
$(x^{2})^{\frac{1}{2}}$
$= x$

The Final Multiplication

Multiply by the outermost term $1 + x^{2}$ :
$1 + x^{2} \cdot x$
$= x 1 + x^{2}$
This matches Option 2.

00:00 / 00:00

The Sigma Insight: Properties of Inverse Trigonometric Functions

Solution Diagram

Analyzing the Setup

The expression

1 + x^{2} [{x cos (cot^{- 1} x) + sin (cot^{- 1} x)}^{2} - 1]^{1/2}

appears complex, but we can simplify it using trigonometric substitution.

Let us define the angle

θ = cot^{- 1} x

. This implies that

cot θ = x

, or more helpfully,

cot θ = \frac{x}{1}

Building the Foundation

Consider a right-angled triangle where

θ

is an acute angle. By the definition of the cotangent function, we set the base to

x

and the perpendicular to

1

Using the Pythagorean theorem, the hypotenuse is calculated as:

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

From this triangle, we derive the following trigonometric ratios:

cos θ = \frac{x}{1 + x ^{2}}, sin θ = \frac{1}{1 + x ^{2}}

Taming the Inner Beast

Now, focus on the inner bracket of the original expression:

x cos θ + sin θ

. Substituting our geometric values, we get:

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

Since

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

, the expression simplifies significantly:

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

The Final Unraveling

We now substitute this result back into the original expression. The term inside the square brackets becomes:

[(1 + x^{2})^{2} - 1]^{1/2} = [1 + x^{2} - 1]^{1/2} = x^{2} = ∣ x ∣

Assuming

x > 0

, the expression simplifies to

1 + x^{2} \cdot x

The final simplified result is $x 1 + x^{2}$ .

Common Traps & FAQs

Wait, why do we assume

cot θ = x /1

? Is there a reason for the denominator?

That is a fantastic question! In trigonometry, we often work with ratios. By writing

cot θ = x

\frac{x}{1}

, we are simply making the ratio explicit:

\frac{Base}{Perpendicular}

. This makes it trivial to identify the base as

x

and the perpendicular as

1

, allowing us to construct our triangle and find the hypotenuse

x^{2} + 1

immediately.

Can I just use trigonometric identities instead of drawing a triangle?

You absolutely can! You could use identities like

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

and

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

. However, drawing the triangle is a powerful, foolproof method that prevents memory errors and helps you visualize the relationship between the sides. It is a tool that never fails, even when you forget the specific identity.

Why does the inner expression simplify so cleanly to

1 + x^{2}

It is not magic, it is the beauty of mathematics! When you substitute

cos θ

and

sin θ

into

x cos θ + sin θ

, you get

\frac{x ^{2}}{1 + x ^{2}} + \frac{1}{1 + x ^{2}}

. Combining these gives

\frac{x ^{2} + 1}{1 + x ^{2}}

. Since

x^{2} + 1

is just

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

, dividing it by

1 + x^{2}

leaves you with exactly

1 + x^{2}

. It is a perfect cancellation designed to reward your persistence.

Animated Solution for Mathematics - Inverse Trigonometric Functions: $1 + x^{2} [{x cos (cot^{- 1} x) + sin (cot^{- 1} x)}^{2} - 1]^{1/2} =$

Select Answer:

Visualized Solution

Defining $θ$

Visualizing the Right Triangle

Calculating the Hypotenuse

Finding $cos θ$

Finding $sin θ$

Substituting into the Inner Expression

Simplifying the Numerator

Simplifying the Fraction

Squaring the Result

Subtracting $1$

Applying the Outer Power

The Final Multiplication

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

Building the Foundation

Taming the Inner Beast

The Final Unraveling

Similar Questions

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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; } 1+x2​[{xcos(cot−1x)+sin(cot−1x)}2−1]1/2=

Select Answer:

Visualized Solution

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

Visualizing the Right Triangle

Calculating the Hypotenuse

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

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

Substituting into the Inner Expression

Simplifying the Numerator

Simplifying the Fraction

Squaring the Result

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

Applying the Outer Power

The Final Multiplication

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

Building the Foundation

Taming the Inner Beast

The Final Unraveling

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Animated Solution for Mathematics - Inverse Trigonometric Functions: $1 + x^{2} [{x cos (cot^{- 1} x) + sin (cot^{- 1} x)}^{2} - 1]^{1/2} =$

Defining $θ$

Finding $cos θ$

Finding $sin θ$

Subtracting $1$

Prove that $cos tan^{- 1} sin cot^{- 1} x = \frac{x ^{2} + 1}{x ^{2} + 2}$ .

Considering the principal values of the inverse trigonometric functions, $sin^{- 1} (\frac{3}{2} x + \frac{1}{2} 1 - x^{2}), - \frac{1}{2} < x < \frac{1}{2}$ , is equal to

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If $\frac{π}{2} \leq x \leq \frac{3 π}{4}$ then $cos^{- 1} (\frac{12}{13} cos x + \frac{5}{13} sin x)$ is equal to

If $sin^{- 1} (x - \frac{x ^{2}}{2} + \frac{x ^{3}}{4} - \dots) + cos^{- 1} (x^{2} - \frac{x ^{4}}{2} + \frac{x ^{6}}{4} - \dots) = \frac{π}{2}$ for $0 < ∣ x ∣ < 2$ , then $x$ equals

If $α > β > γ > 0,$ then the expression $cot^{- 1} {β + \frac{( 1 + β ^{2} )}{( α - β )}} + cot^{- 1} {γ + \frac{( 1 + γ ^{2} )}{( β - γ )}} + cot^{- 1} {α + \frac{( 1 + α ^{2} )}{( γ - α )}}$ is equal to:

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