JEE Main Previous Year Trigonometry Questions and Solutions

Q 21 :

Let $a \leq \tan^{- 1} x + \cot^{- 1} x + \sin^{- 1} x \leq b .$ If $α$ and $β$ denote the minimum and maximum possible values of $a$ and $b$ respectively, then:

$α = 0, β = π$
$α = 0, β = \frac{π}{2}$
$α = \frac{π}{2}, β = π$
$α = - \frac{π}{2}, β = \frac{π}{2}$

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(1)

Ans. $α = 0, β = π$

Explanation:

We know that

$\tan^{- 1} x + \cot^{- 1} x = \frac{π}{2}$

$\tan^{- 1} x + \cot^{- 1} x + \sin^{- 1} x = \frac{π}{2} + \sin^{- 1} x$

$\frac{- π}{2} \leq \sin^{- 1} x \leq \frac{π}{2}$

$\Rightarrow - \frac{π}{2} + \frac{π}{2} \leq \sin^{- 1} x + \frac{π}{2} \leq \frac{π}{2} + \frac{π}{2}$

$\Rightarrow 0 \leq \frac{π}{2} + \sin^{- 1} x \leq π$

$\Rightarrow 0 \leq \tan^{- 1} x + \cot^{- 1} x + \sin^{- 1} x \leq π$

$∴$ $α = 0, β = π$

Q 22 :

The principal value of $\sin^{- 1} (\frac{- \sqrt{3}}{2})$ is:

$- \frac{2 π}{3}$
$- \frac{π}{3}$
$\frac{4 π}{3}$
$\frac{5 π}{3}$

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(2)

Ans. $- \frac{π}{3}$

Explanation:

$\sin^{- 1} (\frac{- \sqrt{3}}{2}) = \sin^{- 1} (- \sin \frac{π}{3})$

$= - \sin^{- 1} (\sin \frac{π}{3}) = - \frac{π}{3}$

Q 23 :

The domain of the function defined by $f (x) = \sin^{- 1} x + \cos x$ is:

$[- 1, 1]$
$[- 1, π + 1]$
$(- \infty, \infty)$
$ϕ$

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(1)

Ans. $[- 1, 1]$

Explanation:

The domain of $\cos x$ is $R$ and the domain of $\sin^{- 1} x$ is $[- 1, 1] .$

$∴$ The domain of $\cos x + \sin^{- 1} x$ is $R \cap [- 1, 1]$ , i.e., $[- 1, 1] .$

Q 24 :

The principal value of $\sec^{- 1} (\frac{- 2}{\sqrt{3}})$ is:

$\frac{5 π}{6}$
$\frac{2 π}{3}$
$\frac{π}{3}$
None of these

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(1)

Ans. $\frac{5 π}{6}$

Explanation:

$\sec^{- 1} (\frac{- 2}{\sqrt{3}}) = π - \sec^{- 1} (\frac{2}{\sqrt{3}})$

$= π - \frac{π}{6} = \frac{5 π}{6}$

Q 25 :

If $\sin^{- 1} x = y,$ then:

$0 \leq y \leq x$
$\frac{- π}{2} \leq y \leq \frac{π}{2}$
$0 < y < π$
$\frac{- π}{2} < y < \frac{π}{2}$

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(2)

Ans. $\frac{- π}{2} \leq y \leq \frac{π}{2}$

Explanation:

Range of $\sin^{- 1} x$ is $[\frac{- π}{2}, \frac{π}{2}]$

$∴$ $\frac{- π}{2} \leq y \leq \frac{π}{2}$

Q 26 :

The domain of the function $\cos^{- 1} (2 x - 1)$ is:

$[0, 1]$
$[- 1, 1]$
$(- 1, 1)$
$[0, π]$

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(1)

Ans. [0, 1]

Explanation:

We know that $\cos^{- 1} x$ is defined for $x \in [- 1, 1]$

$∴$ $f (x) = \cos^{- 1} (2 x - 1)$ is defined if

$- 1 \leq 2 x - 1 \leq 1$

$\Rightarrow$ $0 \leq 2 x \leq 2$

$\Rightarrow$ $0 \leq x \leq 1$

Q 27 :

The value of $\cos^{- 1} (\cos \frac{3 π}{2})$ is equal to:

$\frac{π}{2}$
$\frac{3 π}{2}$
$\frac{5 π}{2}$
$\frac{7 π}{2}$

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(1)

Ans. $\frac{π}{2}$

Explanation:

$\cos^{- 1} (\cos \frac{3 π}{2}) \neq \frac{3 π}{2}$ as $\frac{3 π}{2} \notin [0, π]$

$∴$ $\cos^{- 1} (\cos \frac{3 π}{2}) = \cos^{- 1} (0) = \frac{π}{2}$

Q 28 :

The value of $\sin [2 \sin^{- 1} (0.6)]$ is:

$\cdot 48$
$\cdot 96$
$1 \cdot 2$
$\sin 1 \cdot 2$

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(2)

Ans. $\cdot 96$

Explanation:

Let $\sin^{- 1} (\cdot 6) = θ,$ i.e., $\sin θ = \cdot 6$

Now, $\sin [2 \sin^{- 1} (0.6)] = \sin (2 θ) = 2 \sin θ \cos θ$

$= 2 (\cdot 6) (\cdot 8) = \cdot 96$

$(∵ \cos θ = \sqrt{1 - \sin^{2} θ} = \sqrt{1 - {(0.6)}^{2}} = 0.8)$

Q 29 :

If $\cot^{- 1} (- \frac{1}{5}) = θ$ , the value of $\sin θ$ is:

$\frac{\sqrt{26}}{5}$
$- \frac{5}{\sqrt{26}}$
$\frac{\sqrt{5}}{\sqrt{26}}$
$\frac{5}{\sqrt{26}}$

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(4)

Ans. $\frac{5}{\sqrt{26}}$

Explanation:

Given, $\cot^{- 1} (\frac{- 1}{5}) = θ, where θ \in (0, π)$

$∴$ $\cot θ = \frac{- 1}{5}$

$∵$ $\sin θ = \frac{1}{c o s e c θ} = \frac{1}{\sqrt{1 + \cot^{2} θ}}$

$= \frac{1}{\sqrt{1 + {(- \frac{1}{5})}^{2}}} = \frac{1}{\sqrt{1 + \frac{1}{25}}} = \frac{1}{\sqrt{\frac{26}{25}}}$

$= \frac{5}{\sqrt{26}}$

Q 30 :

$\tan^{- 1} \sqrt{3} - \sec^{- 1} (- 2)$ is equal to:

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

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(2)

Ans. $- \frac{π}{3}$

Explanation:

$\tan^{- 1} \sqrt{3} - \sec^{- 1} (- 2)$

$= \tan^{- 1} \sqrt{3} - (π - \sec^{- 1} 2)$

$= \frac{π}{3} - π + \sec^{- 1} 2$

$= \frac{π}{3} - π + \frac{π}{3}$

$= \frac{2 π}{3} - π$

$= - \frac{π}{3}$

Q 31 :

The value of $\tan [\frac{1}{2} \cos^{- 1} (\frac{\sqrt{5}}{3})]$ is:

$\frac{3 + \sqrt{5}}{2}$
$\frac{3 - \sqrt{5}}{2}$
$\frac{- 3 + \sqrt{5}}{2}$
$\frac{- 3 - \sqrt{5}}{2}$

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(2)

Ans. $\frac{3 - \sqrt{5}}{2}$

Explanation:

Let $y = \tan [\frac{1}{2} \cos^{- 1} (\frac{\sqrt{5}}{3})]$

Putting, $x = \cos^{- 1} (\frac{\sqrt{5}}{3})$

$\Rightarrow$ $\cos x = \frac{\sqrt{5}}{3}$ ...(i)

Now, $y = \tan (\frac{x}{2})$

$\Rightarrow$ $y = \sqrt{\frac{1 - \cos x}{1 + \cos x}}$

$\Rightarrow$ $y = \sqrt{\frac{1 - \frac{\sqrt{5}}{3}}{1 + \frac{\sqrt{5}}{3}}}$ [From (i)]

$\Rightarrow$ $y = \sqrt{\frac{3 - \sqrt{5}}{3 + \sqrt{5}}}$ $= \sqrt{\frac{(3 - \sqrt{5}) (3 - \sqrt{5})}{(3 + \sqrt{5}) (3 - \sqrt{5})}}$

$\Rightarrow$ $y = \sqrt{\frac{{(3 - \sqrt{5})}^{2}}{9 - 5}}$ $= \frac{3 - \sqrt{5}}{\sqrt{4}}$

$\Rightarrow$ $y = \frac{3 - \sqrt{5}}{2}$

i.e., $\tan [\frac{1}{2} \cos^{- 1} (\frac{\sqrt{5}}{3})] = \frac{3 - \sqrt{5}}{2}$

Q 32 :

Match List I with List II.

List I List II

A. The range of $\sin^{- 1} x$ is I. $[\frac{- π}{2}, \frac{π}{2}] - {0}$

B. The range of $\tan^{- 1} x$ is II. $(\frac{π}{2}, \frac{π}{2})$

C. The range of ${cosec}^{- 1} x$ is III. $[0, π] - {\frac{π}{2}}$

D. The range of $\sec^{- 1} x$ is IV. $[\frac{- π}{2}, \frac{π}{2}]$

Choose the correct answer from the option given below.

A-IV, B-III, C-II, D-I
A-IV, B-I, C-III, D-II
A-I, B-IV, C-II, D-III
A-IV, B-II, C-I, D-III

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(4)

Ans. A-IV, B-II, C-I, D-III

Q 33 :

If the equation $x^{3} + b x^{2} + c x + 1 = 0 (b < c)$ has only one real root $α$ . Then the value of $2 \tan^{- 1} (cosec α) + \tan^{- 1} (2 \sin α \sec^{2} α)$ is:

$- \frac{π}{2}$
$- π$
$\frac{π}{2}$
$π$

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(2)

$x^{3} + b x^{2} + c x + 1 = 0$

$f (- 1) = b - c < 0$

$f (0) = 1 > 0$

$B \in (0, 1)$

$y = - 2 \tan^{- 1} (\cos e c B) - \tan^{- 1} (\frac{2 \sin B}{\cos^{2} B})$

$= - (π + \tan^{- 1} \frac{2 \cos B}{1 - c o s e c^{2} B}) - \tan^{- 1} \frac{2 \sin B}{\cos^{2} B}$

$= - π$

Q 34 :

$If \sum_{r = 1}^{n} (\frac{\tan 2^{r - 1}}{\cos 2^{r}}) = \tan p^{n} - \tan q, then find the value of (p + q)$

(3)

$\sum_{r = 1}^{n} \frac{\sin (2^{r} - 2^{r - 1})}{\cos 2^{r} \cos 2^{r - 1}} = \sum_{r = 1}^{n} (\tan 2^{r} - \tan 2^{r - 1}) = \tan 2^{n} - \tan 1$

Q 35 :

Let $p = \tan (\frac{5 π}{9} \cos (2 \sin^{- 1} \frac{1}{\sqrt{5}})),$ $q = \sin^{- 1} (\sin \frac{2 π}{3}) + \cos^{- 1} (\cos (\frac{7 π}{6}))$ then the quadratic equation whose roots are $p$ , $\sec q$ is (Here inverse trigonometric functions take principal values)

$\sqrt{3} x^{2} - x + \sqrt{3} = 0$
$\sqrt{3} x^{2} - 4 x + 2 \sqrt{3} = 0$
$\sqrt{3} x^{2} - x - 2 \sqrt{3} = 0$
$x^{2} - \sqrt{3} x + = 0$

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(3)

$\cos (2 \sin^{- 1} \frac{1}{\sqrt{5}}) = \frac{3}{5},$ $\sin^{- 1} (\sin \frac{2 π}{3}) = \frac{π}{3},$ $\cos^{- 1} (\cos \frac{7 π}{6}) = 2 π - \frac{7 π}{6}$

Q 36 :

Exhaustive value of $x$ such that $\cos^{- 1} (\frac{8 x}{1 + 16 x^{2}}) = - \frac{π}{2} + 2 \tan^{- 1} (4 x)$

$(\frac{1}{8}, \infty)$
$(\frac{1}{16}, \frac{1}{8})$
$[\frac{1}{4}, \infty)$
$(\frac{1}{8}, \frac{1}{4})$

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(3)

$On solving \frac{π}{2} - \sin^{- 1} (\frac{2 \times 4 x}{1 + {(4 x)}^{2}}) = - \frac{π}{2} + 2 \tan^{- 1} (4 x)$

$\Rightarrow 2 \tan^{- 1} (4 x) = π - \sin^{- 1} (\frac{2 (4 x)}{1 + {(4 x)}^{2}})$

$4 x \geq 1; x \geq \frac{1}{4}$

Q 37 :

Which of the following is true?

$\frac{d y}{d x} for y = \sin^{- 1} (\cos x), where x \in (0, π), is - 1$
$\frac{d y}{d x} for y = \sin^{- 1} (\cos x), where x \in (0, 2 π), is 1$
$\frac{d y}{d x} for y = \cos^{- 1} (\sin x), where x \in (- π, \frac{π}{2}), is - 1$
$\frac{d y}{d x} for y = \cos^{- 1} (\sin x), where x \in (- \frac{π}{2}, \frac{3 π}{2}), is - 1$

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(1)

$Use graphs of \sin^{- 1} (\sin x) and \cos^{- 1} (\cos x)$

Q 38 :

$Let S_{k} = \sum_{r = 1}^{k} \tan^{- 1} (\frac{6^{r}}{2^{2 r + 1} + 3^{2 r + 1}}) . Then \lim_{k \to \infty} S_{k} is equal to$

$\tan^{- 1} (\frac{3}{2})$
$\frac{π}{2}$
$\cot^{- 1} (\frac{3}{2})$
$\tan^{- 1} (3)$

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(3)

$S_{k} = \sum_{r = 1}^{k} \tan^{- 1} (\frac{\frac{1}{3} {(\frac{2}{3})}^{r}}{{(\frac{2}{3})}^{2 r + 1} + 1}) = \sum_{r = 1}^{k} [\tan^{- 1} {(\frac{2}{3})}^{r} - \tan^{- 1} {(\frac{2}{3})}^{r + 1}] = \tan^{- 1} {(\frac{2}{3})}^{r} - \tan^{- 1} (0) = \cot^{- 1} (\frac{3}{2})$

Q 39 :

Let $a_{1} = 1, a_{2}, a_{3}, a_{4}, \dots$ be consecutive natural numbers. Then $\tan^{- 1} (\frac{1}{1 + a_{1} a_{2}}) + \tan^{- 1} (\frac{1}{1 + a_{2} a_{3}}) + \dots + \tan^{- 1} (\frac{1}{1 + a_{2024} a_{2025}})$ is equal to

$\frac{π}{4} + \cot^{- 1} (2025)$
$\tan^{- 1} (2025) - \frac{π}{4}$
$\tan^{- 1} (2024) - \frac{π}{4}$
$\frac{π}{4} - \tan^{- 1} (2024)$

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(2)

By given condition, $a_{2} - a_{1} = a_{3} - a_{2} = \dots = a_{2025} - a_{2024} = 1$

$∴$ $\tan^{- 1} (\frac{a_{2} - a_{1}}{1 + a_{1} a_{2}}) + \tan^{- 1} (\frac{a_{3} - a_{2}}{1 + a_{2} a_{3}}) + \dots + \tan^{- 1} (\frac{a_{2025} - a_{2024}}{1 + a_{2024} a_{2025}})$

$[\tan^{- 1} a_{2} - \tan^{- 1} a_{1}] + [\tan^{- 1} a_{3} - \tan^{- 1} a_{2}] + \dots + [\tan^{- 1} a_{2025} - \tan^{- 1} a_{2024}]$

$= \tan^{- 1} a_{2025} - \tan^{- 1} a_{1} = \tan^{- 1} (2025) - \frac{π}{4}$

Q 40 :

Let [.] denote the greatest integer function. If the domain of the function $f (x) = \cos^{- 1} (\frac{4 x + 2 [x]}{3})$ is $[α, β]$ , then $12 (α + β)$ is equal to: [2026]

6
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9
4

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(1)

We have,

$y = \cos^{- 1} (\frac{4 x + 2 [x]}{3})$ , then

$- 1 \leq \frac{4 x + 2 [x]}{3} \leq 1 \Rightarrow - 3 \leq 4 x + 2 [x] \leq 3$

Case 1: If $x \in [0, 1]$ , then $- 3 \leq 4 x \leq 3$

$∴$ $x \in [0, \frac{3}{4}]$

Case 2: If $x \in [1, 2)$ , then $- 3 \leq 4 x + 2 \leq 3 \Rightarrow - 5 \leq 4 x \leq 1$

$∴$ $x \in [\frac{- 5}{4}, \frac{1}{4}]$ $(Reject)$

Case 3: If $x \in [- 1, 0)$ , then $- 3 \leq 4 x - 2 \leq 3 \Rightarrow - 1 \leq 4 x \leq 5$

$\Rightarrow x \in [\frac{- 1}{4}, \frac{5}{4}]$ $∴$ $x \in [\frac{- 1}{4}, 0]$

Case 4: If $x \in [- 2, - 1)$ , then $- 3 \leq 4 x - 4 \leq 3 \Rightarrow 1 \leq 4 x \leq 7$

$∴$ $x \in [\frac{1}{4}, \frac{7}{4}]$ $(Reject)$

$∴$ $Domain : x \in [\frac{- 1}{4}, \frac{3}{4}]$

Hence, $α = \frac{- 1}{4}, β = \frac{3}{4}$

Thus, $12 (α + β) = 12 (\frac{1}{2}) = 6$

Q 41 :

Let $[\cdot]$ denote the greatest integer function. If the domain of the function $f (x) = \sin^{- 1} (\frac{x + [x]}{3})$ is $(α, β)$ , then $α^{2} + β^{2}$ is equal to : [2026]

2
5
10
13

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(2)

$Since, - 1 < \sin^{- 1} θ < 1$

$\Rightarrow - 1 \leq \frac{x + [x]}{3} \leq 1$

$\Rightarrow - 3 \leq x + [x] \leq 3$

$\Rightarrow - 3 - [x] \leq x \leq 3 - [x]$

$If [x] = - 1 \Rightarrow x \in [- 1, 0)$

$If [x] = 0 \Rightarrow x \in [0, 1)$

$If [x] = 1 \Rightarrow x \in [1, 2)$

$Hence, x \in [- 1, 2)$

$\Rightarrow$ $α = - 1 and β = 2$

$So, α^{2} + β^{2} = 5$

Q 42 :

Let $0 < α < 1, β = \frac{1}{3 α}$ and $\tan^{- 1} (1 - α) + \tan^{- 1} (1 - β) = \frac{π}{4}$ . Then $6 (α + β)$ is equal to: [2026]

6
7
8
9

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(2)

We have,

$\tan^{- 1} (1 - α) + \tan^{- 1} (1 - β) = \frac{π}{4}$

$\Rightarrow \tan^{- 1} [\frac{(1 - α) + (1 - β)}{1 - (1 - α) (1 - β)}] = π / 4$

$\Rightarrow \frac{1 - α + 1 - β}{1 - [1 - α - β + α β]} = \tan π / 4$

$\Rightarrow \frac{2 - (α + β)}{α + β - α β} = 1$

$\Rightarrow 2 - (α + β) = (α + β) - α β$

$\Rightarrow 2 (α + β) = 2 + α β$ $\dots (i)$

$∵$ $α β = \frac{1}{3}$ $[∵ Given]$

$∴$ $From (i),$

$α + β = \frac{2 + \frac{1}{3}}{2} \Rightarrow α + β = \frac{7}{6}$

$\Rightarrow 6 (α + β) = 7$

Q 43 :

If $\sin (\tan^{- 1} (x \sqrt{2})) = \cot (\sin^{- 1} \sqrt{1 - x^{2}}), x \in (0, 1)$ , then the value of $x$ is: [2026]

$\frac{1}{2}$
$\frac{1}{3}$
$\frac{2}{3}$
$\frac{5}{8}$

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(1)

Given

$\sin (\tan^{- 1} \sqrt{2} x) = \cot (\sin^{- 1} \sqrt{1 - x^{2}})$

Let $\tan^{- 1} \sqrt{2} x = θ$ and $\sin^{- 1} \sqrt{1 - x^{2}} = ϕ$

$\Rightarrow \sqrt{2} x = \tan θ and \sin ϕ = \sqrt{1 - x^{2}}$

$\Rightarrow \sin θ = \cot ϕ$

$\Rightarrow \frac{\sqrt{2} x}{\sqrt{1 + 2 x^{2}}} = \frac{x}{\sqrt{1 - x^{2}}}$

$\Rightarrow \sqrt{2} \sqrt{1 - x^{2}} = \sqrt{1 + 2 x^{2}}$

Squaring both sides , we get

$2 - 2 x^{2} = 1 + 2 x^{2}$

$\Rightarrow 1 = 4 x^{2}$

$\Rightarrow x = \pm \frac{1}{2}$

$∵$ $x \in (0, 1)$

So, $x = \frac{1}{2}$

Q 44 :

Let $α = 3 \sin^{- 1} (\frac{6}{11})$ and $β = 3 \cos^{- 1} (\frac{4}{9})$ , where inverse trigonometric functions take only the principal values.

Given below are two statements :

Statement I : $\cos (α + β) > 0$ .

Statement II : $\cos (α) < 0$ .

In the light of the above statements, choose the correct answer from the options given below : [2026]

Both Statement I and Statement II are true
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Statement I is false but Statement II is true

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(1)

$\frac{1}{2} < \frac{6}{11} < \frac{1}{\sqrt{2}}$

$\Rightarrow \sin^{- 1} (\frac{1}{2}) < \sin^{- 1} (\frac{6}{11}) < \sin^{- 1} (\frac{1}{\sqrt{2}})$

$\Rightarrow \frac{π}{6} < \sin^{- 1} (\frac{6}{11}) < \frac{π}{4}$

$\Rightarrow \frac{π}{2} < 3 \sin^{- 1} (\frac{6}{11}) < \frac{3 π}{4}$

$\Rightarrow \frac{π}{2} < α < \frac{3 π}{4}$ $∴$ $(\cos α < 0)$ $\dots (i)$

Also, $0 < \frac{4}{9} < \frac{1}{2}$

$\Rightarrow \frac{π}{3} < \cos^{- 1} (\frac{4}{9}) < \frac{π}{2}$

$\Rightarrow π < 3 \cos^{- 1} (\frac{4}{9}) < \frac{3 π}{2}$

$\Rightarrow π < β < \frac{3 π}{2}$ $\dots (ii)$

From (i) and (ii), we get $\frac{3 π}{2} < α + β < \frac{9 π}{4}$ $∴$ $\cos (α + β) > 0$

So, both statement I and II are true.

Q 45 :

If $\frac{π}{4} + \sum_{p = 1}^{11} \tan^{- 1} (\frac{2^{p - 1}}{1 + 2^{2 p - 1}}) = α$ , then $\tan α$ is equal to ______ . [2026]

(2048)

Given,

$\frac{π}{4} + \sum_{p = 1}^{11} \tan^{- 1} (\frac{2^{p - 1}}{1 + 2^{2 p - 1}}) = α$

Let,

$T_{p} = \tan^{- 1} (\frac{2^{p - 1}}{1 + 2^{2 p - 1}}) = \tan^{- 1} (\frac{2^{p} - 2^{p - 1}}{1 + 2^{p} \cdot 2^{p - 1}})$

$= \tan^{- 1} (2^{p}) - \tan^{- 1} (2^{p - 1})$

Now, $T_{1} = \tan^{- 1} (2^{1}) - \tan^{- 1} (2^{0})$

$T_{2} = \tan^{- 1} (2^{2}) - \tan^{- 1} (2^{1})$

$T_{3} = \tan^{- 1} (2^{3}) - \tan^{- 1} (2^{2})$

.................................................

$T_{11} = \tan^{- 1} (2^{11}) - \tan^{- 1} (2^{10})$

$\Rightarrow \sum_{p = 1}^{11} \tan^{- 1} (\frac{2^{p - 1}}{1 + 2^{2 p - 1}}) = \tan^{- 1} (2^{11}) - \frac{π}{4}$

$∴$ $α = \frac{π}{4} + \tan^{- 1} (2^{11}) - \frac{π}{4}$

$\Rightarrow \tan α = 2^{11} \Rightarrow \tan α = 2048$

Topic Question Set

Q 21 :

Let $a \leq \tan^{- 1} x + \cot^{- 1} x + \sin^{- 1} x \leq b .$ If $α$ and $β$ denote the minimum and maximum possible values of $a$ and $b$ respectively, then:

Q 22 :

The principal value of $\sin^{- 1} (\frac{- \sqrt{3}}{2})$ is:

Q 23 :

The domain of the function defined by $f (x) = \sin^{- 1} x + \cos x$ is:

Q 24 :

The principal value of $\sec^{- 1} (\frac{- 2}{\sqrt{3}})$ is:

Q 25 :

If $\sin^{- 1} x = y,$ then:

Q 26 :

The domain of the function $\cos^{- 1} (2 x - 1)$ is:

Q 27 :

The value of $\cos^{- 1} (\cos \frac{3 π}{2})$ is equal to:

Q 28 :

The value of $\sin [2 \sin^{- 1} (0.6)]$ is:

Q 29 :

If $\cot^{- 1} (- \frac{1}{5}) = θ$ , the value of $\sin θ$ is:

Q 30 :

$\tan^{- 1} \sqrt{3} - \sec^{- 1} (- 2)$ is equal to:

Q 31 :

The value of $\tan [\frac{1}{2} \cos^{- 1} (\frac{\sqrt{5}}{3})]$ is:

Q 33 :

If the equation $x^{3} + b x^{2} + c x + 1 = 0 (b < c)$ has only one real root $α$ . Then the value of $2 \tan^{- 1} (cosec α) + \tan^{- 1} (2 \sin α \sec^{2} α)$ is:

Q 34 :

$If \sum_{r = 1}^{n} (\frac{\tan 2^{r - 1}}{\cos 2^{r}}) = \tan p^{n} - \tan q, then find the value of (p + q)$

Q 35 :

Let $p = \tan (\frac{5 π}{9} \cos (2 \sin^{- 1} \frac{1}{\sqrt{5}})),$ $q = \sin^{- 1} (\sin \frac{2 π}{3}) + \cos^{- 1} (\cos (\frac{7 π}{6}))$ then the quadratic equation whose roots are $p$ , $\sec q$ is (Here inverse trigonometric functions take principal values)

Q 36 :

Exhaustive value of $x$ such that $\cos^{- 1} (\frac{8 x}{1 + 16 x^{2}}) = - \frac{π}{2} + 2 \tan^{- 1} (4 x)$

Q 37 :

Which of the following is true?

Q 38 :

$Let S_{k} = \sum_{r = 1}^{k} \tan^{- 1} (\frac{6^{r}}{2^{2 r + 1} + 3^{2 r + 1}}) . Then \lim_{k \to \infty} S_{k} is equal to$

Q 39 :

Let $a_{1} = 1, a_{2}, a_{3}, a_{4}, \dots$ be consecutive natural numbers. Then $\tan^{- 1} (\frac{1}{1 + a_{1} a_{2}}) + \tan^{- 1} (\frac{1}{1 + a_{2} a_{3}}) + \dots + \tan^{- 1} (\frac{1}{1 + a_{2024} a_{2025}})$ is equal to

Q 40 :

Let [.] denote the greatest integer function. If the domain of the function $f (x) = \cos^{- 1} (\frac{4 x + 2 [x]}{3})$ is $[α, β]$ , then $12 (α + β)$ is equal to: [2026]

Q 41 :

Let $[\cdot]$ denote the greatest integer function. If the domain of the function $f (x) = \sin^{- 1} (\frac{x + [x]}{3})$ is $(α, β)$ , then $α^{2} + β^{2}$ is equal to : [2026]

Q 42 :

Let $0 < α < 1, β = \frac{1}{3 α}$ and $\tan^{- 1} (1 - α) + \tan^{- 1} (1 - β) = \frac{π}{4}$ . Then $6 (α + β)$ is equal to: [2026]

Q 43 :

If $\sin (\tan^{- 1} (x \sqrt{2})) = \cot (\sin^{- 1} \sqrt{1 - x^{2}}), x \in (0, 1)$ , then the value of $x$ is: [2026]

Q 45 :

If $\frac{π}{4} + \sum_{p = 1}^{11} \tan^{- 1} (\frac{2^{p - 1}}{1 + 2^{2 p - 1}}) = α$ , then $\tan α$ is equal to ______ . [2026]

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	List I		List II
A.	The range of $\sin^{- 1} x$ is	I.	$[\frac{- π}{2}, \frac{π}{2}] - {0}$
B.	The range of $\tan^{- 1} x$ is	II.	$(\frac{π}{2}, \frac{π}{2})$
C.	The range of ${cosec}^{- 1} x$ is	III.	$[0, π] - {\frac{π}{2}}$
D.	The range of $\sec^{- 1} x$ is	IV.	$[\frac{- π}{2}, \frac{π}{2}]$

Topic Question Set

Q 21 : Let a≤tan-1x+cot-1x+sin-1x≤b. If α and β denote the minimum and maximum possible values of a and b respectively, then:

Q 22 : The principal value of sin-1(-32) is:

Q 23 : The domain of the function defined by f(x)=sin-1x+cosx is:

Q 24 : The principal value of sec-1(-23) is:

Q 25 : If sin-1x=y, then:

Q 26 : The domain of the function cos-1(2x-1) is:

Q 27 : The value of cos-1(cos3π2) is equal to:

Q 28 : The value of sin[2sin-1(0.6)] is:

Q 29 : If cot-1(-15)=θ, the value of sinθ is:

Q 30 : tan-13-sec-1(-2) is equal to:

Q 31 : The value of tan[12cos-1(53)] is:

Q 32 : Match List I with List II. List I List II A. The range of sin-1x is I. [-π2,π2]-{0} B. The range of tan-1x is II. (π2,π2) C. The range of cosec-1x is III. [0,π]-{π2} D. The range of sec-1x is IV. [-π2,π2] Choose the correct answer from the option given below.

Q 33 : If the equation x3+bx2+cx+1=0 (b<c) has only one real root α. Then the value of 2tan-1(cosec α)+tan-1(2sin α sec2 α) is:

Q 34 : If ∑r=1n(tan2r-1cos2r)=tanpn-tanq, then find the value of (p+q)

Q 35 : Let p=tan(5π9cos(2sin-115)), q=sin-1(sin2π3)+cos-1(cos(7π6)) then the quadratic equation whose roots are p, sec q is (Here inverse trigonometric functions take principal values)

Q 36 : Exhaustive value of x such that cos-1(8x1+16x2)=-π2+2tan-1(4x)

Q 37 : Which of the following is true?

Q 38 : Let Sk=∑r=1ktan-1(6r22r+1+32r+1). Then limk→∞Sk is equal to

Q 39 : Let a1=1,a2,a3,a4,… be consecutive natural numbers. Then tan-1(11+a1a2)+tan-1(11+a2a3)+…+tan-1(11+a2024a2025) is equal to

Q 40 : Let [.] denote the greatest integer function. If the domain of the function f(x)=cos-1(4x+2[x]3) is [α,β], then 12(α+β) is equal to: [2026]

Q 41 : Let [·] denote the greatest integer function. If the domain of the function f(x)=sin-1(x+[x]3) is (α,β), then α2+β2 is equal to : [2026]

Q 42 : Let 0<α<1, β=13α and tan-1(1-α)+tan-1(1-β)=π4. Then 6(α+β) is equal to: [2026]

Q 43 : If sin(tan-1(x2))=cot(sin-11-x2), x∈(0,1), then the value of x is: [2026]

Q 45 : If π4+∑p=111tan-1(2p-11+22p-1)=α, then tanα is equal to ______ . [2026]

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Q 21 :

Let $a \leq \tan^{- 1} x + \cot^{- 1} x + \sin^{- 1} x \leq b .$ If $α$ and $β$ denote the minimum and maximum possible values of $a$ and $b$ respectively, then:

Q 22 :

The principal value of $\sin^{- 1} (\frac{- \sqrt{3}}{2})$ is:

Q 23 :

The domain of the function defined by $f (x) = \sin^{- 1} x + \cos x$ is:

Q 24 :

The principal value of $\sec^{- 1} (\frac{- 2}{\sqrt{3}})$ is:

Q 25 :

If $\sin^{- 1} x = y,$ then:

Q 26 :

The domain of the function $\cos^{- 1} (2 x - 1)$ is:

Q 27 :

The value of $\cos^{- 1} (\cos \frac{3 π}{2})$ is equal to:

Q 28 :

The value of $\sin [2 \sin^{- 1} (0.6)]$ is:

Q 29 :

If $\cot^{- 1} (- \frac{1}{5}) = θ$ , the value of $\sin θ$ is:

Q 30 :

$\tan^{- 1} \sqrt{3} - \sec^{- 1} (- 2)$ is equal to:

Q 31 :

The value of $\tan [\frac{1}{2} \cos^{- 1} (\frac{\sqrt{5}}{3})]$ is:

Q 33 :

If the equation $x^{3} + b x^{2} + c x + 1 = 0 (b < c)$ has only one real root $α$ . Then the value of $2 \tan^{- 1} (cosec α) + \tan^{- 1} (2 \sin α \sec^{2} α)$ is:

Q 34 :

$If \sum_{r = 1}^{n} (\frac{\tan 2^{r - 1}}{\cos 2^{r}}) = \tan p^{n} - \tan q, then find the value of (p + q)$

Q 35 :

Let $p = \tan (\frac{5 π}{9} \cos (2 \sin^{- 1} \frac{1}{\sqrt{5}})),$ $q = \sin^{- 1} (\sin \frac{2 π}{3}) + \cos^{- 1} (\cos (\frac{7 π}{6}))$ then the quadratic equation whose roots are $p$ , $\sec q$ is (Here inverse trigonometric functions take principal values)

Q 36 :

Exhaustive value of $x$ such that $\cos^{- 1} (\frac{8 x}{1 + 16 x^{2}}) = - \frac{π}{2} + 2 \tan^{- 1} (4 x)$

Q 37 :

Which of the following is true?

Q 38 :

$Let S_{k} = \sum_{r = 1}^{k} \tan^{- 1} (\frac{6^{r}}{2^{2 r + 1} + 3^{2 r + 1}}) . Then \lim_{k \to \infty} S_{k} is equal to$

Q 39 :

Let $a_{1} = 1, a_{2}, a_{3}, a_{4}, \dots$ be consecutive natural numbers. Then $\tan^{- 1} (\frac{1}{1 + a_{1} a_{2}}) + \tan^{- 1} (\frac{1}{1 + a_{2} a_{3}}) + \dots + \tan^{- 1} (\frac{1}{1 + a_{2024} a_{2025}})$ is equal to

Q 40 :

Let [.] denote the greatest integer function. If the domain of the function $f (x) = \cos^{- 1} (\frac{4 x + 2 [x]}{3})$ is $[α, β]$ , then $12 (α + β)$ is equal to: [2026]

Q 41 :

Let $[\cdot]$ denote the greatest integer function. If the domain of the function $f (x) = \sin^{- 1} (\frac{x + [x]}{3})$ is $(α, β)$ , then $α^{2} + β^{2}$ is equal to : [2026]

Q 42 :

Let $0 < α < 1, β = \frac{1}{3 α}$ and $\tan^{- 1} (1 - α) + \tan^{- 1} (1 - β) = \frac{π}{4}$ . Then $6 (α + β)$ is equal to: [2026]

Q 43 :

If $\sin (\tan^{- 1} (x \sqrt{2})) = \cot (\sin^{- 1} \sqrt{1 - x^{2}}), x \in (0, 1)$ , then the value of $x$ is: [2026]

Q 45 :

If $\frac{π}{4} + \sum_{p = 1}^{11} \tan^{- 1} (\frac{2^{p - 1}}{1 + 2^{2 p - 1}}) = α$ , then $\tan α$ is equal to ______ . [2026]