JEE Advanced Maths PYP Inverse Trigonometric Functions Properties & Infinite Series

Q 1 :

The total number of real solutions of the equation $θ = \tan^{- 1} (2 \tan θ) - \frac{1}{2} \sin^{- 1} (\frac{6 \tan θ}{9 + \tan^{2} θ})$ is

(Here, the inverse trigonometric functions $\sin^{- 1} x$ and $\tan^{- 1} x$ assume values in $[- \frac{π}{2}, \frac{π}{2}]$ and $(- \frac{π}{2}, \frac{π}{2})$ , respectively.) [2025]

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5

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(3)

$θ = \tan^{- 1} (2 \tan θ) - \frac{1}{2} \sin^{- 1} (\frac{6 \tan θ}{9 + \tan^{2} θ})$

Let $\tan θ = 1$ and taking tangent on both sides,

$t = \frac{2 t - \tan (\frac{1}{2} \sin^{- 1} (\frac{6 t}{9 + t^{2}}))}{1 + 2 t \tan (\frac{1}{2} \sin^{- 1} (\frac{6 t}{9 + t^{2}}))}$ $\dots (i)$

Let $\sin^{- 1} (\frac{6 t}{9 + t^{2}}) = α$

$\sin α = \frac{6 t}{9 + t^{2}}$

$\frac{2 \tan \frac{α}{2}}{1 + \tan^{2} \frac{α}{2}} = \frac{2 (\frac{t}{3})}{1 + {(\frac{t}{3})}^{2}} = \frac{2 (\frac{3}{t})}{{(\frac{3}{t})}^{2} + 1}$

So, $\tan \frac{α}{2} = \frac{3}{t} or \frac{t}{3}$

$t = \frac{2 t - \tan (\frac{α}{2})}{1 + 2 t . \tan (\frac{α}{2})}$ [From (i)]

Case-I: If $\tan \frac{α}{2} = \frac{3}{t}$

$\Rightarrow t = \frac{2 t - \frac{3}{t}}{1 + 6}$ $\Rightarrow 7 t^{2} = 2 t^{2} - 3$

$\Rightarrow 5 t^{2} = - 3,$

which is not possible.

Case-II: If $\tan \frac{α}{2} = \frac{t}{3},$

$\Rightarrow t = \frac{2 t - \frac{t}{3}}{1 + \frac{2 t^{2}}{3}}$ $\Rightarrow t (3 + 2 t^{2}) = 5 t$

$\Rightarrow 3 + 2 t^{2} = 5$

$\Rightarrow t = 0, 1, - 1 = \tan θ$

$\Rightarrow θ = 0, \frac{π}{4}, - \frac{π}{4}$

Therefore, the solution are ${\frac{- π}{4}, 0, \frac{π}{4}}$

Q 2 :

The value of $\cot (\sum_{n = 1}^{23} \cot^{- 1} (1 + \sum_{k = 1}^{n} 2 k))$ is [2013]

$\frac{23}{25}$
$\frac{25}{23}$
$\frac{23}{24}$
$\frac{24}{23}$

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(2)

$\cot^{- 1} (1 + \sum_{k = 1}^{n} 2 k) = \cot^{- 1} [1 + n (n + 1)]$

$= \tan^{- 1} (\frac{(n + 1) - n}{1 + (n + 1) n}) = \tan^{- 1} (n + 1) - \tan^{- 1} n$

$∴$ $\sum_{n = 1}^{23} [\tan^{- 1} (n + 1) - \tan^{- 1} n] = \tan^{- 1} 24 - \tan^{- 1} 1 = \tan^{- 1} \frac{23}{25}$

$∴$ $\cot [\sum_{n = 1}^{23} \cot^{- 1} (1 + \sum_{k = 1}^{n} 2 k)] = \cot [\tan^{- 1} \frac{23}{25}] = \frac{25}{23}$

Q 3 :

If $0 < x < 1$ , then $\sqrt{1 + x^{2}} {[{x \cos (\cot^{- 1} x) + \sin (\cot^{- 1} x)}^{2} - 1]}^{1 / 2} =$ [2008]

$\frac{x}{\sqrt{1 + x^{2}}}$
$x$
$x \sqrt{1 + x^{2}}$
$\sqrt{1 + x^{2}}$

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(3)

$\sqrt{1 + x^{2}} {[{x \cos (\cot^{- 1} x) + \sin (\cot^{- 1} x)}^{2} - 1]}^{\frac{1}{2}}$

$= \sqrt{1 + x^{2}} {[{x \cos (\cos^{- 1} (\frac{x}{\sqrt{1 + x^{2}}})) + \sin (\sin^{- 1} (\frac{1}{\sqrt{1 + x^{2}}}))}^{2} - 1]}^{\frac{1}{2}}$

$= \sqrt{1 + x^{2}} {[{x \cdot \frac{x}{\sqrt{1 + x^{2}}} + \frac{1}{\sqrt{1 + x^{2}}}}^{2} - 1]}^{\frac{1}{2}}$

$= \sqrt{1 + x^{2}} {[{(\sqrt{1 + x^{2}})}^{2} - 1]}^{\frac{1}{2}}$

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

Q 4 :

The value of $x$ for which $\sin (\cot^{- 1} (1 + x)) = \cos (\tan^{- 1} x)$ is [2004]

$\frac{1}{2}$
$1$
$0$
$- \frac{1}{2}$

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(4)

$\sin [\cot^{- 1} (1 + x)] = \cos (\tan^{- 1} x)$

$\Rightarrow \sin [\sin^{- 1} (\frac{1}{\sqrt{1 + {(1 + x)}^{2}}})] = \cos [\cos^{- 1} (\frac{1}{\sqrt{1 + x^{2}}})]$

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

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

$\Rightarrow 2 x + 1 = 0$

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

Q 5 :

If $\sin^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \dots) + \cos^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - \dots) = \frac{π}{2}$ for $0 < | x | < \sqrt{2}$ , then $x$ equals [2001]

$\frac{1}{2}$
$1$
$- \frac{1}{2}$
$- 1$

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(2)

$\sin^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \dots) + \cos^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - \dots) = \frac{π}{2}$

$\Rightarrow \cos^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - \dots) = \frac{π}{2} - \sin^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \dots)$

$\Rightarrow \cos^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - \dots) = \cos^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \dots)$

$\Rightarrow x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - \dots = x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \dots$

On both sides we have G.P. infinite terms.

$∴$ $\frac{x^{2}}{1 - (- \frac{x^{2}}{2})} = \frac{x}{1 - (- \frac{x}{2})}$ $\Rightarrow \frac{2 x^{2}}{2 + x^{2}} = \frac{2 x}{2 + x}$

$\Rightarrow 2 x + x^{3} = 2 x^{2} + x^{3}$ $\Rightarrow x (x - 1) = 0$

$\Rightarrow x = 0, 1$ but $0 < | x | < \sqrt{2} \Rightarrow x = 1$

Q 6 :

Let $\tan^{- 1} (x) \in (- \frac{π}{2}, \frac{π}{2})$ for $x \in ℝ$ . Then the number of real solutions of the equation $\sqrt{1 + \cos (2 x)} = \sqrt{2} \tan^{- 1} (\tan x)$ in the set

$(- \frac{3 π}{2}, - \frac{π}{2}) \cup (- \frac{π}{2}, \frac{π}{2}) \cup (\frac{π}{2}, \frac{3 π}{2})$ is equal to [2023]

(3)

$\sqrt{1 + \cos 2 x} = \sqrt{2} \tan^{- 1} (\tan x)$

$\Rightarrow \sqrt{2 \cos^{2} x} = \sqrt{2} \tan^{- 1} (\tan x)$

$\Rightarrow | \cos x | = \tan^{- 1} (\tan x)$

[IMAGE 1211]

$Number of solutions = Number of intersection points = 3$

Q 7 :

The number of real solutions of the equation $\sin^{- 1} (\sum_{i = 1}^{\infty} x^{i + 1} - x \sum_{i = 1}^{\infty} {(\frac{x}{2})}^{i}) = \frac{π}{2} - \cos^{- 1} (\sum_{i = 1}^{\infty} {(- \frac{x}{2})}^{i} - \sum_{i = 1}^{\infty} {(- x)}^{i})$ lying in the interval $(- \frac{1}{2}, \frac{1}{2})$ is ______.

(Here, the inverse trigonometric functions $\sin^{- 1} x$ and $\cos^{- 1} x$ assume values in $[- \frac{π}{2}, \frac{π}{2}]$ and $[0, π],$ respectively.) [2018]

(2)

$\sin^{- 1} (\sum_{i = 1}^{\infty} x^{i + 1} - {x \sum}_{i = 1}^{\infty} {(\frac{x}{2})}^{i})$

$= \frac{π}{2} - \cos^{- 1} [\sum_{i = 1}^{\infty} {(- \frac{x}{2})}^{i} - \sum_{i = 1}^{\infty} {(- x)}^{i}]$

$= \sin^{- 1} (\frac{x^{2}}{1 - x} - x \cdot \frac{\frac{x}{2}}{1 - \frac{x}{2}}) = \sin^{- 1} (\frac{- \frac{x}{2}}{1 + \frac{x}{2}} - \frac{- x}{1 + x})$

$[∵ Sum of infinite terms of a G.P. = \frac{a}{1 - r}, if | r | < 1]$

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

$\Rightarrow \frac{x^{2}}{1 - x} - \frac{x}{1 + x} + \frac{x}{2 + x} - \frac{x^{2}}{2 - x} = 0$

$\Rightarrow \frac{x (x^{2} + 2 x - 1)}{1 - x^{2}} + \frac{x (2 - 3 x - x^{2})}{4 - x^{2}} = 0$

$\Rightarrow x (x^{3} + 2 x^{2} + 5 x - 2) = 0$

$\Rightarrow x = 0 or x^{3} + 2 x^{2} + 5 x - 2 = 0 = p (x) (say)$

$We observe that p (0) < 0 and p (\frac{1}{2}) > 0$

$∴$ $One root of p (x) = 0 lies in (0, \frac{1}{2})$

$∴$ $Two solutions lie between - \frac{1}{2} and \frac{1}{2}$

Q 8 :

Considering only the principal values of the inverse trigonometric functions, the value of

$\frac{3}{2} \cos^{- 1} \sqrt{\frac{2}{2 + π^{2}}} + \frac{1}{4} \sin^{- 1} (\frac{2 \sqrt{2} π}{2 + π^{2}}) + \tan^{- 1} (\frac{\sqrt{2}}{π})$ is ________. [2022]

(2.36)

$[Let, \cos^{- 1} \sqrt{\frac{2}{2 + π^{2}}} = t = \tan^{- 1} (\frac{π}{\sqrt{2}})]$

${Similarly for \sin^{- 1} (\frac{2 \sqrt{2} π}{2 + π^{2}})}$

$Now, we have$

$\frac{3}{2} \tan^{- 1} (\frac{π}{\sqrt{2}}) + \frac{1}{4} \tan^{- 1} (\frac{2 \sqrt{2} π}{π^{2} - 2}) + \tan^{- 1} (\frac{\sqrt{2}}{π})$

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

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

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

$= \frac{π}{2} + \frac{π}{4} = \frac{3 π}{4} = 2.36$

Q 9 :

For any $y \in ℝ$ , let $\cot^{- 1} (y) \in (0, π)$ and $\tan^{- 1} (y) \in (- \frac{π}{2}, \frac{π}{2}) .$ Then the sum of all solutions of the equation

$\tan^{- 1} (\frac{6 y}{9 - y^{2}}) + \cot^{- 1} (\frac{9 - y^{2}}{6 y}) = \frac{2 π}{3},$ for $0 < | y | < 3$ , is equal to [2023]

$2 \sqrt{3} - 3$
$3 - 2 \sqrt{3}$
$4 \sqrt{3} - 6$
$6 - 4 \sqrt{3}$

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(3)

$Case-I: y \in (- 3, 0) \Rightarrow y < 0 \Rightarrow \frac{6 y}{9 - y^{2}} < 0$

$\tan^{- 1} (\frac{6 y}{9 - y^{2}}) + π + \tan^{- 1} (\frac{6 y}{9 - y^{2}}) = \frac{2 π}{3}$

$2 \tan^{- 1} (\frac{6 y}{9 - y^{2}}) = - \frac{π}{3}$

$y^{2} - 6 \sqrt{3} y - 9 = 0 \Rightarrow y = 3 \sqrt{3} - 6$ $(∵ y \in (- 3, 0))$

$Case-II: y \in (0, 3) \Rightarrow y > 0 \Rightarrow \frac{6 y}{9 - y^{2}} > 0$

$2 \tan^{- 1} (\frac{6 y}{9 - y^{2}}) = \frac{2 π}{3} \Rightarrow \sqrt{3} y^{2} + 6 y - 9 \sqrt{3} = 0$

$y = \sqrt{3} or y = - 3 \sqrt{3}$ $(rejected)$

$∴$ $Sum = \sqrt{3} + (3 \sqrt{3} - 6) = 4 \sqrt{3} - 6$

Q 10 :

For any positive integer $n$ , let $S_{n} : (0, \infty) \to ℝ$ be defined by $S_{n} (x) = \sum_{k = 1}^{n} \cot^{- 1} (\frac{1 + k (k + 1) x^{2}}{x}),$ where for any $x \in ℝ$ ,
$\cot^{- 1} x \in (0, π)$ and $\tan^{- 1} (x) \in (- \frac{π}{2}, \frac{π}{2}) .$ Then which of the following statements is (are) TRUE? [2021]

$S_{10} (x) = \frac{π}{2} - \tan^{- 1} (\frac{1 + 11 x^{2}}{10 x}), for all x > 0$
$\lim_{n \to \infty} \cot (S_{n} (x)) = x, for all x > 0$
The equation $S_{3} (x) = \frac{π}{4}$ has a root in $(0, \infty)$
$\tan (S_{n} (x)) \leq \frac{1}{2}, for all n \geq 1 and x > 0$

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Select one or more options

(1, 2)

$Given that S_{n} (x) = \sum_{k = 1}^{n} \cot^{- 1} (\frac{1 + k (k + 1) x^{2}}{x})$

$= \sum_{k = 1}^{n} \tan^{- 1} (\frac{x}{1 + k x (k x + x)})$

$= \sum_{k = 1}^{n} \tan^{- 1} (\frac{(k x + x) - (k x)}{1 + (k x + x) (k x)})$

$\Rightarrow S_{n} (x) = \tan^{- 1} (n x + x) - \tan^{- 1} x$ $= \tan^{- 1} (\frac{n x}{1 + (n + 1) x^{2}})$

$(1) S_{10} (x) = \tan^{- 1} (\frac{10 x}{1 + 11 x^{2}}) = \frac{π}{2} - \tan^{- 1} (\frac{1 + 11 x^{2}}{10 x}), (x > 0)$

$Option (1) is correct$

$(2) \lim_{n \to \infty} \cot (S_{n} (x)) = \lim_{n \to \infty} \cot [\cot^{- 1} (\frac{1 + (n + 1) x^{2}}{n x})]$

$= \lim_{n \to \infty} \frac{\frac{1}{n} + (1 + \frac{1}{n}) x^{2}}{x} = x, (x > 0)$

$Option (2) is correct$

$(3) S_{3} (x) = \tan^{- 1} (\frac{3 x}{1 + 4 x^{2}}) = \frac{π}{4}$

$\Rightarrow 4 x^{2} - 3 x + 1 = 0 \Rightarrow x \notin ℝ$ $[∵ D is negative]$

$Option (3) is incorrect$

$(4) For x = 1,$

$\tan (S_{n} (x)) = \frac{n}{n + 2} \geq \frac{1}{2}, for n \geq 3$

$Option (4) is incorrect$

Q 11 :

For non-negative integers $n$ , let $f (n) = \frac{\sum_{k = 0}^{n} \sin (\frac{k + 1}{n + 2} π) \sin (\frac{k + 2}{n + 2} π)}{\sum_{k = 0}^{n} \sin^{2} (\frac{k + 1}{n + 2} π)} .$

Assuming $\cos^{- 1} x$ takes values in $[0, π]$ , which of the following options is/are correct? [2019]

$\lim_{n \to \infty} f (n) = \frac{1}{2}$
$f (4) = \frac{\sqrt{3}}{2}$
If $α = \tan (\cos^{- 1} f (6)),$ then $α^{2} + 2 α - 1 = 0$
$\sin (7 \cos^{- 1} f (5)) = 0$

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Select one or more options

(2, 3, 4)

$f (n) = \frac{\sum_{k = 0}^{n} 2 \sin (\frac{k + 1}{n + 2} π) \sin (\frac{k + 2}{n + 2} π)}{\sum_{k = 0}^{n} 2 \sin^{2} (\frac{k + 1}{n + 2}) π},$

where $n$ is a non-negative integer.

$= \frac{\sum_{k = 0}^{n} [\cos (\frac{π}{n + 2}) - \cos (\frac{(2 k + 3) π}{n + 2})]}{\sum_{k = 0}^{n} [1 - \cos (\frac{2 (k + 1) π}{n + 2})]}$

$= \frac{(n + 1) \cos (\frac{π}{n + 2}) - [\cos (\frac{3 π}{n + 2}) + \cos (\frac{5 π}{n + 2}) + \dots + \cos (\frac{(2 n + 3) π}{n + 2})]}{n + 1 - [\cos (\frac{2 π}{n + 2}) + \cos (\frac{4 π}{n + 2}) + \dots + \cos (\frac{2 (n + 1) π}{n + 2})]}$

$= \frac{(n + 1) \cos (\frac{π}{n + 2}) - \frac{\sin (\frac{(n + 1) π}{n + 2})}{\sin (\frac{π}{n + 2})} \cos (\frac{(2 n + 6) π}{2 (n + 2)})}{n + 1 - \frac{\sin (\frac{(n + 1) π}{n + 2})}{\sin (\frac{π}{n + 2})} \cos (\frac{(2 n + 4) π}{2 (n + 2)})}$

$= \frac{(n + 1) \cos (\frac{π}{n + 2}) + \cos (\frac{π}{n + 2})}{(n + 1) + 1} = \frac{(n + 2) \cos (\frac{π}{n + 2})}{n + 2}$

$∴$ $f (n) = \cos (\frac{π}{n + 2})$

$\lim_{n \to \infty} f (n) = \lim_{n \to \infty} \cos (\frac{π}{n + 2}) = 1$

$∴$ $Option (1) is incorrect$

$f (4) = \cos (\frac{π}{4 + 2}) = \cos (\frac{π}{6}) = \frac{\sqrt{3}}{2}$

$∴$ $Option (2) is correct$

$If α = \tan (\cos^{- 1} f (6)),$

$= \tan (\cos^{- 1} (\cos \frac{π}{8})) = \tan \frac{π}{8}$

Now, $\tan \frac{π}{4} = 1 = \frac{2 \tan \frac{π}{8}}{1 - \tan^{2} \frac{π}{8}}$

$\Rightarrow \frac{2 α}{1 - α^{2}} = 1 \Rightarrow α^{2} + 2 α - 1 = 0$

$∴$ $Option (3) is correct$

$\sin (7 \cos^{- 1} f (5)) = \sin (7 \cos^{- 1} (\cos \frac{π}{7})) = \sin (7 \times \frac{π}{7})$

$= \sin π = 0$

$∴$ $Option (4) is correct$

Q 12 :

Let $(x, y)$ be such that $\sin^{- 1} (a x) + \cos^{- 1} (y) + \cos^{- 1} (b x y) = \frac{π}{2} .$

Match the statements in Column I with statements in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS. [2007]

Column I Column II

(A) If $a = 1$ and $b = 0$ , then $(x, y)$ (p) lies on the circle $x^{2} + y^{2} = 1$

(B) If $a = 1$ and $b = 1$ , then $(x, y)$ (q) lies on $(x^{2} - 1) (y^{2} - 1) = 0$

(C) If $a = 1$ and $b = 2$ , then $(x, y)$ (r) lies on $y = x$

(D) If $a = 2$ and $b = 2$ , then $(x, y)$ (s) lies on $(4 x^{2} - 1) (y^{2} - 1) = 0$

(A) → s; (B) → p; (C) → q; (D) → q
(A) → s; (B) → p; (C) → q; (D) → p
(A) → s; (B) → q; (C) → p; (D) → p
(A) → p; (B) → q; (C) → p; (D) → s

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(4)

(A) → p; (B) → q; (C) → p; (D) → s

$\sin^{- 1} (a x) + \cos^{- 1} y + \cos^{- 1} (b x y) = \frac{π}{2}$

$\Rightarrow \cos^{- 1} y + \cos^{- 1} (b x y) = \frac{π}{2} - \sin^{- 1} (a x) = \cos^{- 1} (a x)$

$Let \cos^{- 1} y = α, \cos^{- 1} (b x y) = β, \cos^{- 1} (a x) = γ$

$Then y = \cos α, b x y = \cos β, a x = \cos γ$

$∴$ we get $α + β = γ$ and $\cos β = b x y$

$\Rightarrow \cos (γ - α) = \cos β = b x y$

$\Rightarrow \cos γ \cos α + \sin γ \sin α = b x y$

$\Rightarrow a x y + \sin γ \sin α = b x y \Rightarrow (a - b) x y = - \sin α \sin γ$

$\Rightarrow {(a - b)}^{2} x^{2} y^{2} = \sin^{2} α \sin^{2} γ$

$= (1 - \cos^{2} α) (1 - \cos^{2} γ)$

$∴$ ${(a - b)}^{2} x^{2} y^{2} = (1 - y^{2}) (1 - a^{2} x^{2})$ $\dots (1)$

$(A)$ $For a = 1, b = 0, equation (i) reduces to$

$x^{2} y^{2} = (1 - x^{2}) (1 - y^{2}) \Rightarrow x^{2} + y^{2} = 1$

$(B)$ $For a = 1, b = 1, equation (i) becomes$

$(1 - x^{2}) (1 - y^{2}) = 0 \Rightarrow (x^{2} - 1) (y^{2} - 1) = 0$

$(C)$ $For a = 1, b = 2, equation (i) reduces to$

$x^{2} y^{2} = (1 - x^{2}) (1 - y^{2}) \Rightarrow x^{2} + y^{2} = 1$

$(D)$ $For a = 2, b = 2, equation (i) reduces to$

$0 = (1 - 4 x^{2}) (1 - y^{2}) \Rightarrow (4 x^{2} - 1) (y^{2} - 1) = 0$

Q 13 :

Match the following. [2006]

Column I Column II

(A) $\sum_{i = 1}^{\infty} \tan^{- 1} (\frac{1}{2 i^{2}}) = t, then \tan t =$ (p) $1$

(B) Sides $a, b, c$ of a triangle $A B C$ are in A.P. and
$\cos θ_{1} = \frac{a}{b + c}, \cos θ_{2} = \frac{b}{a + c}, \cos θ_{3} = \frac{c}{a + b},$
then $\tan^{2} (\frac{θ_{1}}{2}) + \tan^{2} (\frac{θ_{3}}{2}) =$ (q) $\frac{\sqrt{5}}{3}$

(C) A line is perpendicular to $x + 2 y + 2 z = 0$
and passes through $(0, 1, 0) .$ The perpendicular
distance of this line from the origin is (r) $\frac{2}{3}$

(A) → (p); (B) → (r); (C) → (q)
(A) → (r); (B) →(p); (C) →(q)
(A) → (q); (B) →(p); (C) → (r)
(A) → (q); (B) → (r); (C) → (p)

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(1)

(A) → (p); (B) → (r); (C) → (q)

$(A)$ $t = \sum_{i = 1}^{\infty} \tan^{- 1} (\frac{1}{2 i^{2}}) = \sum_{i = 1}^{\infty} \tan^{- 1} [\frac{(2 i + 1) - (2 i - 1)}{1 + 4 i^{2} - 1}]$

$= \sum_{i = 1}^{\infty} [\tan^{- 1} (2 i + 1) - \tan^{- 1} (2 i - 1)]$

$= \tan^{- 1} 3 - \tan^{- 1} 1 + \tan^{- 1} 5 - \tan^{- 1} 3 + \dots + \tan^{- 1} (2 n + 1) - \tan^{- 1} (2 n - 1) + \dots \infty$

$= \lim_{n \to \infty} [\tan^{- 1} (2 n + 1) - \tan^{- 1} 1]$

$= \lim_{n \to \infty} \tan^{- 1} [\frac{2 n}{1 + (2 n + 1)}] = \lim_{n \to \infty} \tan^{- 1} (\frac{1}{1 + \frac{1}{n}})$

$= \tan^{- 1} (1) = \frac{π}{4} \Rightarrow \tan t = 1,$ $(A) \to (p)$

$(B)$ $∵$ $a, b, c are in A.P. \Rightarrow 2 b = a + c$

$Now \cos θ_{1} = \frac{a}{b + c} \Rightarrow \frac{1 - \tan^{2} \frac{θ_{1}}{2}}{1 + \tan^{2} \frac{θ_{1}}{2}} = \frac{a}{b + c}$

$\Rightarrow \tan^{2} \frac{θ_{1}}{2} = \frac{b + c - a}{b + c + a}$

$Similarly, \tan^{2} \frac{θ_{3}}{2} = \frac{a + b - c}{a + b + c}$

$\Rightarrow \tan^{2} \frac{θ_{1}}{2} + \tan^{2} \frac{θ_{3}}{2} = \frac{2 b}{a + b + c} = \frac{2 b}{3 b} = \frac{2}{3},$ $(B) \to (r)$

$(C)$ $Equation of line through (0, 1, 0) and perpendicular to x + 2 y + 2 z = 0 is$

$\frac{x}{1} = \frac{y - 1}{2} = \frac{z}{2} = λ$

$For some value of λ, the foot of perpendicular from origin to line is (λ, 2 λ + 1, 2 λ)$

$Direction ratios of this perpendicular from origin are λ, 2 λ + 1, 2 λ$

$∴$ $1 \cdot λ + 2 (2 λ + 1) + 2 \cdot 2 λ = 0 \Rightarrow λ = - \frac{2}{9}$

$∴$ $Foot of perpendicular is (- \frac{2}{9}, \frac{5}{9}, - \frac{4}{9})$

$Hence required distance = \sqrt{\frac{4}{81} + \frac{25}{81} + \frac{16}{81}} = \sqrt{\frac{45}{81}} = \frac{\sqrt{5}}{3},$ $(C) \to (q)$

Topic Question Set

Q 2 :

The value of $\cot (\sum_{n = 1}^{23} \cot^{- 1} (1 + \sum_{k = 1}^{n} 2 k))$ is [2013]

Q 3 :

If $0 < x < 1$ , then $\sqrt{1 + x^{2}} {[{x \cos (\cot^{- 1} x) + \sin (\cot^{- 1} x)}^{2} - 1]}^{1 / 2} =$ [2008]

Q 4 :

The value of $x$ for which $\sin (\cot^{- 1} (1 + x)) = \cos (\tan^{- 1} x)$ is [2004]

Q 5 :

If $\sin^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \dots) + \cos^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - \dots) = \frac{π}{2}$ for $0 < | x | < \sqrt{2}$ , then $x$ equals [2001]

Q 8 :

Considering only the principal values of the inverse trigonometric functions, the value of

$\frac{3}{2} \cos^{- 1} \sqrt{\frac{2}{2 + π^{2}}} + \frac{1}{4} \sin^{- 1} (\frac{2 \sqrt{2} π}{2 + π^{2}}) + \tan^{- 1} (\frac{\sqrt{2}}{π})$ is ________. [2022]

Q 9 :

For any $y \in ℝ$ , let $\cot^{- 1} (y) \in (0, π)$ and $\tan^{- 1} (y) \in (- \frac{π}{2}, \frac{π}{2}) .$ Then the sum of all solutions of the equation

$\tan^{- 1} (\frac{6 y}{9 - y^{2}}) + \cot^{- 1} (\frac{9 - y^{2}}{6 y}) = \frac{2 π}{3},$ for $0 < | y | < 3$ , is equal to [2023]

Q 11 :

For non-negative integers $n$ , let $f (n) = \frac{\sum_{k = 0}^{n} \sin (\frac{k + 1}{n + 2} π) \sin (\frac{k + 2}{n + 2} π)}{\sum_{k = 0}^{n} \sin^{2} (\frac{k + 1}{n + 2} π)} .$

Assuming $\cos^{- 1} x$ takes values in $[0, π]$ , which of the following options is/are correct? [2019]

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	Column I		Column II
(A)	If $a = 1$ and $b = 0$ , then $(x, y)$	(p)	lies on the circle $x^{2} + y^{2} = 1$
(B)	If $a = 1$ and $b = 1$ , then $(x, y)$	(q)	lies on $(x^{2} - 1) (y^{2} - 1) = 0$
(C)	If $a = 1$ and $b = 2$ , then $(x, y)$	(r)	lies on $y = x$
(D)	If $a = 2$ and $b = 2$ , then $(x, y)$	(s)	lies on $(4 x^{2} - 1) (y^{2} - 1) = 0$

	Column I		Column II
(A)	$\sum_{i = 1}^{\infty} \tan^{- 1} (\frac{1}{2 i^{2}}) = t, then \tan t =$	(p)	$1$
(B)	Sides $a, b, c$ of a triangle $A B C$ are in A.P. and $\cos θ_{1} = \frac{a}{b + c}, \cos θ_{2} = \frac{b}{a + c}, \cos θ_{3} = \frac{c}{a + b},$ then $\tan^{2} (\frac{θ_{1}}{2}) + \tan^{2} (\frac{θ_{3}}{2}) =$	(q)	$\frac{\sqrt{5}}{3}$
(C)	A line is perpendicular to $x + 2 y + 2 z = 0$ and passes through $(0, 1, 0) .$ The perpendicular distance of this line from the origin is	(r)	$\frac{2}{3}$

Topic Question Set

Q 1 : The total number of real solutions of the equation θ=tan-1(2tanθ)-12sin-1(6tanθ9+tan2θ) is (Here, the inverse trigonometric functions sin-1x and tan-1x assume values in [-π2,π2] and (-π2,π2), respectively.) [2025]

Q 2 : The value of cot(∑n=123cot-1(1+∑k=1n2k)) is [2013]

Q 3 : If 0<x<1, then 1+x2[{xcos(cot-1x)+sin(cot-1x)}2-1]1/2= [2008]

Q 4 : The value of x for which sin(cot-1(1+x))=cos(tan-1x) is [2004]

Q 5 : If sin-1(x-x22+x34-⋯)+cos-1(x2-x42+x64-⋯)=π2 for 0<|x|<2, then x equals [2001]

Q 6 : Let tan-1(x)∈(-π2,π2) for x∈ℝ. Then the number of real solutions of the equation 1+cos(2x)=2tan-1(tanx) in the set (-3π2,-π2)∪(-π2,π2)∪(π2,3π2) is equal to [2023]

Q 7 : The number of real solutions of the equation sin-1(∑i=1∞x i+1-x∑i=1∞(x2)i)=π2-cos-1(∑i=1∞(-x2)i-∑i=1∞(-x)i) lying in the interval (-12,12) is ______. (Here, the inverse trigonometric functions sin-1x and cos-1x assume values in [-π2,π2] and [0,π], respectively.) [2018]

Q 8 : Considering only the principal values of the inverse trigonometric functions, the value of 32cos-122+π2+14sin-1(22π2+π2)+tan-1(2π) is ________. [2022]

Q 9 : For any y∈ℝ, let cot-1(y)∈(0,π) and tan-1(y)∈(-π2,π2). Then the sum of all solutions of the equation tan-1(6y9-y2)+cot-1(9-y26y)=2π3, for 0<|y|<3, is equal to [2023]

Q 10 : For any positive integer n, let Sn:(0,∞)→ℝ be defined by Sn(x)=∑k=1ncot-1(1+k(k+1)x2x), where for any x∈ℝ, cot-1x∈(0,π) and tan-1(x)∈(-π2,π2). Then which of the following statements is (are) TRUE? [2021]

Q 11 : For non-negative integers n, let f(n)=∑k=0nsin(k+1n+2π)sin(k+2n+2π)∑k=0nsin2(k+1n+2π). Assuming cos-1x takes values in [0,π], which of the following options is/are correct? [2019]

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Q 2 :

The value of $\cot (\sum_{n = 1}^{23} \cot^{- 1} (1 + \sum_{k = 1}^{n} 2 k))$ is [2013]

Q 3 :

If $0 < x < 1$ , then $\sqrt{1 + x^{2}} {[{x \cos (\cot^{- 1} x) + \sin (\cot^{- 1} x)}^{2} - 1]}^{1 / 2} =$ [2008]

Q 4 :

The value of $x$ for which $\sin (\cot^{- 1} (1 + x)) = \cos (\tan^{- 1} x)$ is [2004]

Q 5 :

If $\sin^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - \dots) + \cos^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - \dots) = \frac{π}{2}$ for $0 < | x | < \sqrt{2}$ , then $x$ equals [2001]

Q 8 :

Considering only the principal values of the inverse trigonometric functions, the value of

$\frac{3}{2} \cos^{- 1} \sqrt{\frac{2}{2 + π^{2}}} + \frac{1}{4} \sin^{- 1} (\frac{2 \sqrt{2} π}{2 + π^{2}}) + \tan^{- 1} (\frac{\sqrt{2}}{π})$ is ________. [2022]

Q 9 :

For any $y \in ℝ$ , let $\cot^{- 1} (y) \in (0, π)$ and $\tan^{- 1} (y) \in (- \frac{π}{2}, \frac{π}{2}) .$ Then the sum of all solutions of the equation

$\tan^{- 1} (\frac{6 y}{9 - y^{2}}) + \cot^{- 1} (\frac{9 - y^{2}}{6 y}) = \frac{2 π}{3},$ for $0 < | y | < 3$ , is equal to [2023]

Q 11 :

For non-negative integers $n$ , let $f (n) = \frac{\sum_{k = 0}^{n} \sin (\frac{k + 1}{n + 2} π) \sin (\frac{k + 2}{n + 2} π)}{\sum_{k = 0}^{n} \sin^{2} (\frac{k + 1}{n + 2} π)} .$

Assuming $\cos^{- 1} x$ takes values in $[0, π]$ , which of the following options is/are correct? [2019]