Inverse Trigonometric Functions — JEE Main PYQs

96 previous year questions on Inverse Trigonometric Functions (Mathematics), newest first. Open any question for its complete step-by-step solution.

$tan (2 tan^{- 1} \frac{1}{5} + sec^{- 1} \frac{5}{2} + 2 tan^{- 1} \frac{1}{8})$ is equal to:
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The value of $lim_{n \to \infty} 6 tan {\sum_{r = 1}^{n} tan^{- 1} (\frac{1}{r ^{2} + 3 r + 3})}$ is equal to
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The value of $cot (\sum_{n = 1}^{50} tan^{- 1} (\frac{1}{1 + n + n ^{2}}))$ is
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If the inverse trigonometric functions take principal values, then $cos^{- 1} (\frac{3}{10} cos (tan^{- 1} (\frac{4}{3})) + \frac{2}{5} sin (tan^{- 1} (\frac{4}{3})))$ is equal to :
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The domain of the function $cos^{- 1} (\frac{2 s i n ^{- 1} ( \frac{1}{4 x ^{2} - 1} )}{π})$ is :
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$50 tan (3 tan^{- 1} (\frac{1}{2}) + 2 cos^{- 1} (\frac{1}{5})) + 42 tan (\frac{1}{2} tan^{- 1} (22))$ is equal to ____.
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Let $f (x) = 2 cos^{- 1} x + 4 cot^{- 1} x - 3 x^{2} - 2 x + 10, x \in [- 1, 1]$ . If $[a, b]$ is the range of the function then $4 a - b$ is equal to:
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$sin^{- 1} (sin \frac{2 π}{3}) + cos^{- 1} (cos \frac{7 π}{6}) + tan^{- 1} (tan \frac{3 π}{4})$ is equal to :
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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 :
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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
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The value of $tan^{- 1} (\frac{c o s ( \frac{15 π}{4} ) - 1}{s i n ( \frac{π}{4} )})$ is equal to
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Let $f (x) = cos (2 tan^{- 1} sin (cot^{- 1} \frac{1 - x}{x})), 0 < x < 1$ . Then:
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$lim_{n \to \infty} tan {\sum_{r = 1}^{n} tan^{- 1} (\frac{1}{1 + r + r ^{2}})}$ is equal to _____.
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The domain of the function $cosec^{- 1} (\frac{1 + x}{x})$ is:
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If $\sum_{r = 1}^{50} tan^{- 1} \frac{1}{2 r ^{2}} = p$ , then the value of $tan p$ is:
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The number of real roots of the equation $tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = \frac{π}{4}$ is :
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A possible value of $tan (\frac{1}{4} sin^{- 1} \frac{63}{8})$ is:
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The value of $tan (2 tan^{- 1} (\frac{3}{5}) + sin^{- 1} (\frac{5}{13}))$ is equal to :
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If the domain of the function $f (x) = \frac{c o s ^{- 1} x ^{2} - x + 1}{s i n ^{- 1} ( \frac{2 x - 1}{2} )}$ is the interval $(α, β]$ , then $α + β$ is equal to:
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If $\frac{s i n ^{- 1} x}{a} = \frac{c o s ^{- 1} x}{b} = \frac{t a n ^{- 1} y}{c}; 0 < x < 1$ , then the value of $cos (\frac{π c}{a + b})$ is :
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If $(sin^{- 1} x)^{2} - (cos^{- 1} x)^{2} = a, 0 < x < 1, a \neq = 0$ , then the value of $2 x^{2} - 1$ is
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If $0 < a, b < 1$ , and $tan^{- 1} a + tan^{- 1} b = \frac{π}{4}$ , then the value of $(a + b) - (\frac{a ^{2} + b ^{2}}{2}) + (\frac{a ^{3} + b ^{3}}{3}) - (\frac{a ^{4} + b ^{4}}{4}) + \dots\dots\dots\dots\dots$ is
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The domain of the function $f (x) = sin^{- 1} (\frac{3 x ^{2} + x - 1}{( x - 1 ) ^{2}}) + cos^{- 1} (\frac{x - 1}{x + 1})$ is :
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If $cot^{- 1} (α) = cot^{- 1} 2 + cot^{- 1} 8 + cot^{- 1} 18 + cot^{- 1} 32 + \dots$ upto $100$ terms, then $α$ is:
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The sum of possible values of $x$ for $tan^{- 1} (x + 1) + cot^{- 1} (\frac{1}{x - 1}) = tan^{- 1} (\frac{8}{31})$ is :
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$cos^{- 1} (cos (- 5)) + sin^{- 1} (sin (6)) - tan^{- 1} (tan (12))$ is equal to:
(The inverse trigonometric functions take the principal values)
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Let $S_{k} = \sum_{r = 1}^{k} tan^{- 1} (\frac{6 ^{r}}{2 ^{2 r + 1} + 3 ^{2 r + 1}})$ . Then $lim_{k \to \infty} S_{k}$ is equal to :
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Given that the inverse trigonometric functions take principal values only. Then, the number of real values of $x$ which satisfy $sin^{- 1} (\frac{3 x}{5}) + sin^{- 1} (\frac{4 x}{5}) = sin^{- 1} x$ is equal to :
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The number of solutions of the equation $sin^{- 1} [x^{2} + \frac{1}{3}] + cos^{- 1} [x^{2} - \frac{2}{3}] = x^{2}$ , for $x \in [- 1, 1]$ and $[x]$ denotes the greatest less than or equal to $x$ , is:
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Let $f (x) = (sin (tan^{- 1} x) + sin (cot^{- 1} x))^{2} - 1, ∣ x ∣ > 1$ . If $\frac{dy}{dx} = \frac{1}{2} \frac{d}{dx} (sin^{- 1} (f (x)))$ and …
JEE Main 2020View solution →
$2 π - (sin^{- 1} \frac{4}{5} + sin^{- 1} \frac{5}{13} + sin^{- 1} \frac{16}{65})$ is equal to:
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If S is the sum of the first 10 terms of the series $tan^{- 1} (\frac{1}{3}) + tan^{- 1} (\frac{1}{7}) + tan^{- 1} (\frac{1}{13}) + tan^{- 1} (\frac{1}{21}) + \dots .$ , then $tan (S)$ is equal to:
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If $α = cos^{- 1} (\frac{3}{5}), β = tan^{- 1} (\frac{1}{3})$ , where $0 < α, β < \frac{π}{2}$ , then $α - β$ is equal to:
JEE Main 2019View solution →
All $x$ satisfying the inequality $(cot^{- 1} x)^{2} - 7 (cot^{- 1} x) + 10 > 0$ , lie in the interval:
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The value of $sin^{- 1} (\frac{12}{13}) - sin^{- 1} (\frac{3}{5})$ is equal to :
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If $cos^{- 1} x - cos^{- 1} \frac{y}{2} = α$ , where $- 1 \leq x \leq 1, - 2 \leq y \leq 2, x \leq \frac{y}{2}$ , then for all $x, y, 4 x^{2} - 4 x y cos α + y^{2}$ is equal to
JEE Main 2019View solution →
Considering only the principal values of inverse functions, the set $A = {x \geq 0 : tan^{- 1} (2 x) + tan^{- 1} (3 x) = \frac{π}{4}}$
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If $x = sin^{- 1} (sin 10)$ and $y = cos^{- 1} (cos 10)$ , then $y - x$ is equal to:
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If $cos^{- 1} (\frac{2}{3 x}) + cos^{- 1} (\frac{3}{4 x}) = \frac{π}{2}, x > \frac{3}{4}$ then $x$ is equal to:
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Let $tan^{- 1} y = tan^{- 1} x + tan^{- 1} (\frac{2 x}{1 - x ^{2}})$ , where $∣ x ∣ < \frac{1}{3}$ . Then a value of y is :
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If $x, y, z$ are in A.P. and $tan^{- 1} x, tan^{- 1} y$ and $tan^{- 1} z$ are also in A.P., then
JEE Main 2013View solution →
The value of $cot (cosec^{- 1} \frac{5}{3} + tan^{- 1} \frac{2}{3})$ is
JEE Main 2008View solution →
If $sin^{- 1} (\frac{x}{5}) + cosec^{- 1} (\frac{5}{4}) = \frac{π}{2}$ then a value of $x$ is
JEE Main 2007View solution →
Let $f : (- 1, 1) \to B$ , be a function defined by $f (x) = tan^{- 1} \frac{2 x}{1 - x ^{2}}$ , then $f$ is both one-one and onto when $B$ is the interval
JEE Main 2005View solution →
If $cos^{- 1} x - cos^{- 1} \frac{y}{2} = α$ , then $4 x^{2} - 4 x y cos α + y^{2}$ is equal to
JEE Main 2005View solution →
The domain of the function $f (x) = \frac{s i n ^{- 1} ( x - 3 )}{9 - x ^{2}}$ is
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