New chat PYQ Library Doubts Rank report

Sign in to see Recents

Sign in to save progress →

Inverse Trigonometric Functions question

2023 · 30 Jan · Shift 2 · Q23

Guest · try one question or challengeGuest · try one question or challenge

PYQ Library
JEE Main
Mathematics
Inverse Trigonometric Functions
2023 · 30 Jan · Shift 2 · Q23

Inverse Trigonometric Functions question

2023 · 30 Jan · Shift 2 · Q23

JEE MainMathematicsInverse Trigonometric FunctionsMultiple correct+4 / −1

Let

a_{1}=1, a_{2}, a_{3}, a_{4}, \ldots .

. be consecutive natural numbers. Then

\tan ^{-1}\left(\frac{1}{1+a_{1} a_{2}}\right)+\tan ^{-1}\left(\frac{1}{1+a_{2} a_{3}}\right)+\ldots . .+\tan ^{-1}\left(\frac{1}{1+a_{2021} a_{2022}}\right)

is equal to :

A
$\frac{\pi}{4}-\cot ^{-1}(2022)$
B
$\frac{\pi}{4}-\tan ^{-1}(2022)$
C
$\cot ^{-1}(2022)-\frac{\pi}{4}$
D
$\tan ^{-1}(2022)-\frac{\pi}{4}$

More from Inverse Trigonometric Functions

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 :2023 · MCQ
Let (a, b) $\subset (0, 2 π)$ be the largest interval for which $sin^{- 1} (sin θ) - cos^{- 1} (sin θ) > 0, θ \in (0, 2 π)$ , holds. If $α x^{2} + β x + sin^{- 1} (x^{2} - 6 x + 10) + cos^{- 1} (x^{2} - 6 x + 10) = 0$ …2023 · MCQ
The set of all values of k for which $(tan^{- 1} x)^{3} + (cot^{- 1} x)^{3} = k π^{3}, x \in R$ , is the interval :2022 · MCQ
The domain of the function $f (x) = \frac{c o s ^{- 1} ( \frac{x ^{2} - 5 x + 6}{x ^{2} - 9} )}{l o g _{e} ( x ^{2} - 3 x + 2 )}$ is :2022 · MCQ
Let $x * y = x^{2} + y^{3}$ and $(x * 1) * 1 = x * (1 * 1)$ . Then a value of $2 sin^{- 1} (\frac{x ^{4} + x ^{2} - 2}{x ^{4} + x ^{2} + 2})$ is :2022 · MCQ
Let $x = sin (2 tan^{- 1} α)$ and $y = sin (\frac{1}{2} tan^{- 1} \frac{4}{3})$ . If $S = {a \in R : y^{2} = 1 - x}$ , then $α \in S \sum 16 α^{3}$ is equal to …2022 · Numerical
The value of $tan^{- 1} (\frac{c o s ( \frac{15 π}{4} ) - 1}{s i n ( \frac{π}{4} )})$ is equal to :2022 · MCQ
$tan (2 tan^{- 1} \frac{1}{5} + sec^{- 1} \frac{5}{2} + 2 tan^{- 1} \frac{1}{8})$ is equal to :2022 · MCQ