The value of $\cot ^{-1}\left(\frac{\sqrt{1+\tan ^{2}(2)}-1}{\tan (2)}\right)-\cot ^{-1} \left(\frac{\sqrt{1+\tan ^{2}\left(\frac{1}{2}\right)}+1}{\tan \left(\frac{1}{2}\right)}\right)$ is equal to

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Inverse Trigonometric Functions

The value of

cot^{- 1} (\frac{1 + t a n ^{2} ( 2 ) - 1}{t a n ( 2 )}) - cot^{- 1} (\frac{1 + t a n ^{2} ( \frac{1}{2} ) + 1}{t a n ( \frac{1}{2} )})

is equal to

Answer: (A)

Using

1 + tan^{2} θ = ∣ sec θ ∣

: for the first term,

2 \in (\frac{π}{2}, π)

sec 2 < 0

and

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

, giving

cot^{- 1} (- cot 1) = π - 1

. For the second term,

\frac{1}{2} \in (0, \frac{π}{2})

sec \frac{1}{2} > 0

and

\frac{∣ sec \frac{1}{2} ∣ + 1}{tan \frac{1}{2}} = \frac{1 + cos \frac{1}{2}}{sin \frac{1}{2}} = cot \frac{1}{4}

, giving

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

. The expression equals

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

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Single correct option