JEE Main 2025 (Online) 8th April Evening Shift | Inverse Trigonometric Functions Question 1 | Mathematics

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

$ \pi - \frac{3}{2} $

$ \pi + \frac{5}{2} $

$ \pi - \frac{5}{4} $

$ \pi + \frac{3}{2} $

JEE Main 2025 (Online) 4th April Evening Shift

MCQ (Single Correct Answer)

-1

The sum of the infinite series $\cot ^{-1}\left(\frac{7}{4}\right)+\cot ^{-1}\left(\frac{19}{4}\right)+\cot ^{-1}\left(\frac{39}{4}\right)+\cot ^{-1}\left(\frac{67}{4}\right)+\ldots$. is :

$\frac{\pi}{2}+\cot ^{-1}\left(\frac{1}{2}\right)$

$\frac{\pi}{2}-\cot ^{-1}\left(\frac{1}{2}\right)$

$\frac{\pi}{2}-\tan ^{-1}\left(\frac{1}{2}\right)$

$\frac{\pi}{2}+\tan ^{-1}\left(\frac{1}{2}\right)$

JEE Main 2025 (Online) 4th April Morning Shift

MCQ (Single Correct Answer)

-1

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

$\frac{-5 \pi}{6}-\sin ^{-1} x$

$\frac{5 \pi}{6}-\sin ^{-1} x$

$\frac{\pi}{6}+\sin ^{-1} x$

$\frac{\pi}{4}+\sin ^{-1} x$

JEE Main 2025 (Online) 28th January Evening Shift

MCQ (Single Correct Answer)

-1

Let [x] denote the greatest integer less than or equal to x. Then the domain of $ f(x) = \sec^{-1}(2[x] + 1) $ is:

$(-\infty, \infty)$

$(-\infty, \infty)- \{0\}$

$(-\infty, -1] \cup [0, \infty)$

$(-\infty, -1] \cup [1, \infty)$