The value of cot^(-1)((sqrt(1+tan^2(2))-1)/tan(2))-cot^(-1)((sqrt(1+tan^2(1/2))+1)/tan(1/2)) is equal to

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q13 – JEE-MAIN 2025

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Home Maths Inverse TrigonometryQuestion #234

The value of

\cot^{- 1} (\frac{\sqrt{1 + \tan^{2} (2)} - 1}{\tan (2)}) - \cot^{- 1} (\frac{\sqrt{1 + \tan^{2} (\frac{1}{2})} + 1}{\tan (\frac{1}{2})})

is equal to

Clue 1 of 3: Simplify the terms

Use the identity $\sqrt{1 + \tan^{2} x} = | \sec x |$ and consider the quadrant of the angle $x$ .

🥷

Ninja Strategy•Range Analysis

By analyzing the range of each inverse cotangent term, it can be determined that the overall expression must be less than $π$ , which eliminates options that are greater than $π$ .

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Step 1: Simplify the first term✦ Active

Let the first term be $A = \cot^{- 1} (\frac{\sqrt{1 + \tan^{2} (2)} - 1}{\tan (2)})$ . Since $2$ radians is in the second quadrant (approx ${114.6}^{\circ}$ ), $\tan (2) < 0$ and $\sec (2) < 0$ . Thus, $\sqrt{1 + \tan^{2} (2)} = | \sec (2) | = - \sec (2)$ . Substituting this, we get:

A = \cot^{- 1} (\frac{- \sec (2) - 1}{\tan (2)}) = \cot^{- 1} (\frac{- 1 / \cos (2) - 1}{\sin (2) / \cos (2)}) = \cot^{- 1} (\frac{- (1 + \cos (2))}{\sin (2)})

Using half-angle identities

1 + \cos (2) = 2 \cos^{2} (1)

and

\sin (2) = 2 \sin (1) \cos (1)

A = \cot^{- 1} (\frac{- 2 \cos^{2} (1)}{2 \sin (1) \cos (1)}) = \cot^{- 1} (- \cot (1))

Since

1 \in (0, π)

, we have

\cot^{- 1} (- \cot (1)) = π - \cot^{- 1} (\cot (1)) = π - 1

Step 2: Simplify the second term○ Expand

Let the second term be $B = \cot^{- 1} (\frac{\sqrt{1 + \tan^{2} (\frac{1}{2})} + 1}{\tan (\frac{1}{2})})$ . Since $\frac{1}{2}$ radians is in the first quadrant (approx ${28.65}^{\circ}$ ), $\tan (\frac{1}{2}) > 0$ and $\sec (\frac{1}{2}) > 0$ . Thus, $\sqrt{1 + \tan^{2} (\frac{1}{2})} = | \sec (\frac{1}{2}) | = \sec (\frac{1}{2})$ . Substituting this, we get:

B = \cot^{- 1} (\frac{\sec (\frac{1}{2}) + 1}{\tan (\frac{1}{2})}) = \cot^{- 1} (\frac{1 / \cos (\frac{1}{2}) + 1}{\sin (\frac{1}{2}) / \cos (\frac{1}{2})}) = \cot^{- 1} (\frac{1 + \cos (\frac{1}{2})}{\sin (\frac{1}{2})})

Using half-angle identities

1 + \cos (\frac{1}{2}) = 2 \cos^{2} (\frac{1}{4})

and

\sin (\frac{1}{2}) = 2 \sin (\frac{1}{4}) \cos (\frac{1}{4})

B = \cot^{- 1} (\frac{2 \cos^{2} (\frac{1}{4})}{2 \sin (\frac{1}{4}) \cos (\frac{1}{4})}) = \cot^{- 1} (\cot (\frac{1}{4}))

Since

\frac{1}{4} \in (0, π)

, we have

\cot^{- 1} (\cot (\frac{1}{4})) = \frac{1}{4}

Step 3: Calculate the final value○ Expand

The given expression is the difference between the first and second terms, $A - B$ :

A - B = (π - 1) - \frac{1}{4} = π - 1 - \frac{1}{4} = π - \frac{4}{4} - \frac{1}{4} = π - \frac{5}{4}

💡 Teacher's Secret Hint

Ensure correct handling of signs for trigonometric functions in different quadrants.

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Maths Question 13 – JEE-MAIN 2025

Clue 1 of 3: Simplify the terms

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