JEE Main Previous Year Trigonometry Questions and Solutions

Q 11 :

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 $α x^{2} + β x + \sin^{- 1} (x^{2} - 6 x + 10) + \cos^{- 1} (x^{2} - 6 x + 10) = 0$ and $α - β = b - a,$ then $α$ is equal to [2023]

$\frac{π}{8}$
$\frac{π}{48}$
$\frac{π}{16}$
$\frac{π}{12}$

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(4)

We know that $\sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}$

$\sin^{- 1} (\sin θ) - (\frac{π}{2} - \sin^{- 1} (\sin θ)) > 0;$ This equation holds true if, $\sin^{- 1} (\sin θ) > \frac{π}{4} \Rightarrow \sin θ > \frac{1}{\sqrt{2}}$ So, $θ \in (\frac{π}{4}, \frac{3 π}{4})$

$θ \in (\frac{π}{4}, \frac{3 π}{4}) = (a, b) \Rightarrow b - a = \frac{π}{2} = α - β [Given]$

$\Rightarrow β = α - \frac{π}{2}$

$Now, α x^{2} + β x + \sin^{- 1} [{(x - 3)}^{2} + 1] + \cos^{- 1} [{(x - 3)}^{2} + 1] = 0$

$\sin^{- 1} [{(x - 3)}^{2} + 1] + \cos^{- 1} [{(x - 3)}^{2} + 1] = \frac{π}{2} if x = 3$

$Put x = 3, 9 α + 3 β + \frac{π}{2} = 0; Put β = α - \frac{π}{2}$

$\Rightarrow 9 α + 3 (α - \frac{π}{2}) + \frac{π}{2} = 0$

$\Rightarrow 12 α - π = 0 \Rightarrow α = \frac{π}{12}$

Q 12 :

If the domain of the function $f (x) = \sec^{- 1} (\frac{2 x}{5 x + 3})$ is $[α, β) \cup (γ, δ],$ then $| 3 α + 10 (β + γ) + 21 δ |$ is equal to ______ . [2023]

(24)

$Given f (x) = \sec^{- 1} (\frac{2 x}{5 x + 3})$

$So, \frac{2 x}{5 x + 3} \geq 1$

$\Rightarrow \frac{2 x}{5 x + 3} - 1 \geq 0 \Rightarrow \frac{- 3 x - 3}{5 x + 3} \geq 0$

$\Rightarrow \frac{x + 1}{5 x + 3} \leq 0 \Rightarrow x \in [- 1, \frac{- 3}{5})$

$Now, \frac{2 x}{5 x + 3} \leq - 1 \Rightarrow \frac{2 x}{5 x + 3} + 1 \leq 0$

$\Rightarrow \frac{7 x + 3}{5 x + 3} \leq 0 \Rightarrow x \in (\frac{- 3}{5}, \frac{- 3}{7}]$

$∴$ $x \in [- 1, \frac{- 3}{5}) \cup (\frac{- 3}{5}, \frac{- 3}{7}]$

$α = - 1,$ $β = \frac{- 3}{5},$ $γ = \frac{- 3}{5},$ $δ = \frac{- 3}{7}$

$So,$ $| 3 α + 10 (β + γ) + 21 δ |$

$=$ $| 3 \times (- 1) + 10 (\frac{- 3}{5} - \frac{3}{5}) + 21 \times (\frac{- 3}{7}) |$ $=$ $| - 3 - 12 - 9 |$ $= 24$

Q 13 :

If $S = {x \in ℝ : \sin^{- 1} (\frac{x + 1}{\sqrt{x^{2} + 2 x + 2}}) - \sin^{- 1} (\frac{x}{\sqrt{x^{2} + 1}}) = \frac{π}{4}},$ then $\sum_{x \in S} (\sin ((x^{2} + x + 5) \frac{π}{2}) - \cos ((x^{2} + x + 5) π))$ is equal to _____ . [2023]

(4)

$Given, \sin^{- 1} (\frac{x + 1}{\sqrt{x^{2} + 2 x + 2}}) - \sin^{- 1} (\frac{x}{\sqrt{x^{2} + 1}}) = \frac{π}{4}$

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

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

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

$\Rightarrow x = 0 or - 1$ $∴$ $S = {- 1, 0}$

$∴$ $\sum_{x \in S} [\sin ((x^{2} + x + 5) \frac{π}{2}) - \cos ((x^{2} + x + 5) π)]$

$= [\sin (\frac{5 π}{2}) - \cos (5 π)] + {\sin (\frac{5 π}{2}) - \cos (5 π)} = 2 + 2 = 4$

Q 14 :

For $x \in (- 1, 1],$ the number of solutions of the equation $\sin^{- 1} x = 2 \tan^{- 1} x$ is equal to ________ . [2023]

(2)

$\sin^{- 1} x = 2 \tan^{- 1} x$

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

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

$\Rightarrow x = 0 or x^{2} - 1 = 0 \Rightarrow x^{2} = 1 \Rightarrow x = \pm 1$ $\Rightarrow x = 0 or x = 1, - 1$

Number of solutions = 2

$∵$ $x = - 1 \notin (- 1, 1]$

Q 15 :

If the sum of all the solutions of $\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{π}{3}, - 1 < x < 1, x \neq 0,$ is $α - \frac{4}{\sqrt{3}},$ then $α$ is equal to _______ . [2023]

(2)

$Given, \tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{π}{3}$

$Case I : If x > 0$

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

$\Rightarrow \frac{2 x}{1 - x^{2}} = \frac{1}{\sqrt{3}} \Rightarrow x^{2} + 2 \sqrt{3} x - 1 = 0$

$\Rightarrow [x - (2 - \sqrt{3})] [x + (2 + \sqrt{3})] = 0 \Rightarrow x = 2 - \sqrt{3}, - (2 + \sqrt{3})$

$x = - (2 + \sqrt{3}) is rejected because x > 0 . Hence, x = 2 - \sqrt{3}$

$Case II : If x < 0$

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

$\Rightarrow \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = - \frac{π}{3} \Rightarrow \frac{2 x}{1 - x^{2}} = - \sqrt{3}$

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

$\Rightarrow x = \frac{- 1}{\sqrt{3}}, \sqrt{3} \Rightarrow x = \sqrt{3} is rejected because x < 0 .$

$Thus x = - \frac{1}{\sqrt{3}}$

$∴$ $Sum of solutions = 2 - \sqrt{3} - \frac{1}{\sqrt{3}} = 2 - \frac{4}{\sqrt{3}}$

$α - \frac{4}{\sqrt{3}} = 2 - \frac{4}{\sqrt{3}} \Rightarrow α = 2$

Q 16 :

If the domain of the function $f (x)$ $= \cos^{- 1} (\frac{2 x - 5}{11 - 3 x}) + \sin^{- 1} (2 x^{2} - 3 x + 1)$ is the interval $[α, β]$ then $α + 2 β$ is equal to: [2026]

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(3)

$f (x)$ $= \cos^{- 1} (\frac{2 x - 5}{11 - 3 x}) + \sin^{- 1} (2 x^{2} - 3 x + 1)$

$- 1 \leq \frac{2 x - 5}{11 - 3 x} \leq 1$

$- 1 \leq 2 x^{2} - 3 x + 1 \leq 1$

$2 x^{2} - 3 x + 2 \geq 0, 2 x^{2} - 3 x \leq 0$

$x \in [0, \frac{3}{2}] . . . . . . . . . . (i)$

$\frac{2 x - 5}{11 - 3 x} + 1 \geq 0 \frac{2 x - 5}{11 - 3 x} - 1 \leq 0$

$\frac{2 x - 5 + 11 - 3 x}{11 - 3 x} \geq 0 \frac{5 x - 16}{11 - 3 x} \leq 0$

$\frac{6 - x}{11 - 3 x} \geq 0$

$x \in (- \infty, \frac{16}{5}] \cup (\frac{11}{3}, \infty)$

$x \in (- \infty, \frac{11}{3}) \cup (6, \infty)$

Intersection:

$x \in (- \infty, \frac{16}{5}] \cup [6, \infty) . . . . (ii)$

Intersection of (i) & (ii) $x \in [0, \frac{3}{2}]$

$α = 0, β = \frac{3}{2} \Rightarrow α + 2 β = 3$

Q 17 :

The number of solutions of $\tan^{- 1} 4 x + \tan^{- 1} 6 x = \frac{π}{6},$ where $- \frac{1}{2 \sqrt{6}} < x < \frac{1}{2 \sqrt{6}},$ is equal to [2026]

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(2)

$\tan^{- 1} (4 x) + \tan^{- 1} (6 x) = \frac{π}{6}$

$\Rightarrow \tan^{- 1} (\frac{4 x + 6 x}{1 + 24 x^{2}}) = \frac{π}{6}$

$\Rightarrow \frac{10 x}{1 - 24 x^{2}} = \frac{1}{\sqrt{3}}$

$\Rightarrow 24 x^{2} + 10 \sqrt{3} x - 1 = 0$

$x = \frac{- 10 \sqrt{3} \pm \sqrt{300 + 96}}{48}$

$x = \frac{\sqrt{396} - 10 \sqrt{3}}{48}$

$Only 1 solution in (- \frac{1}{2 \sqrt{6}}, \frac{1}{2 \sqrt{6}})$

Q 18 :

Considering the principal values of inverse trigonometric functions, the value of the expression $\tan (2 \sin^{- 1} (\frac{2}{\sqrt{13}}) - 2 \cos^{- 1} (\frac{3}{\sqrt{10}}))$ is equal to: [2026]

$\frac{33}{56}$
$\frac{16}{63}$
$- \frac{33}{56}$
$- \frac{16}{63}$

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(1)

$Let \sin^{- 1} (\frac{2}{\sqrt{13}}) = θ & \cos^{- 1} (\frac{3}{\sqrt{10}}) = ϕ$

$\sin θ = \frac{2}{\sqrt{13}} & \cos ϕ = \frac{3}{\sqrt{10}}$

$\tan (2 θ - 2 ϕ) = \frac{\tan 2 θ - \tan 2 ϕ}{1 + \tan 2 θ \tan 2 ϕ}$

$(∵ \tan 2 θ = \frac{2 \tan θ}{1 - \tan^{2} θ})$

$= \frac{\frac{12}{5} - \frac{3}{4}}{1 + \frac{12}{5} \cdot \frac{3}{4}}$

$= \frac{33}{56}$

Q 19 :

If the domain of the function $f (x)$ $= \sin^{- 1} (\frac{1}{x^{2} - 2 x - 2})$ , is $(- \infty, α] \cup [β, γ] \cup [δ, \infty),$ then $α + β + γ + δ$ is equal to [2026]

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(1)

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

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

$\Rightarrow \frac{(x - 1 - \sqrt{2}) (x - 1 + \sqrt{2})}{(x - 1 - \sqrt{3}) (x - 1 + \sqrt{3})} \geq 0$

$x \in (- \infty, 1 - \sqrt{3}) \cup [1 - \sqrt{2}, 1 + \sqrt{2}] \cup (1 + \sqrt{3}, 0) . . . (1)$

$1 - \frac{1}{x^{2} - 2 x - 2} \geq 0 \Rightarrow \frac{x^{2} - 2 x - 3}{x^{2} - 2 x - 2} \geq 0$

$\Rightarrow \frac{(x + 1) (x - 3)}{(x - 1 - \sqrt{3}) (x - 1 + \sqrt{3})} \geq 0$

$x \in (- \infty, - 1] \cup (1 - \sqrt{3}, 1 + \sqrt{3}) \cup [3, \infty) . . . (2)$

$(1) \cap (2)$

$\Rightarrow x \in (- \infty, - 1] \cup [1 - \sqrt{2}, 1 + \sqrt{2}] \cup [3, \infty)$

$∴$ $α + β + γ + δ = 4$

Q 20 :

Let the maximum value of ${(\sin^{- 1} x)}^{2} + {(\cos^{- 1} x)}^{2}$ for $x \in [- \frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}]$ be $\frac{m}{n} π^{2}$ , where $\gcd (m, n) = 1$ . Then $m + n$ is equal to_____ [2026]

(65)

${(\sin^{- 1} x)}^{2} + {(\cos^{- 1} x)}^{2}$

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

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

$= 2 {(\sin^{- 1} x - \frac{π}{4})}^{2} + \frac{π^{2}}{8}, where \sin^{- 1} x \in [\frac{- π}{3}, \frac{π}{4}]$

Then maximum value occurs at $\sin^{- 1} x = \frac{- π}{3}$

$Which is 2 {(\frac{π}{3} + \frac{π}{4})}^{2} + \frac{π^{2}}{8} = \frac{29 π^{2}}{36}$

$\Rightarrow m = 29 and n = 36$

$∴$ $m + n = 65$

Q 21 :

If $k = \tan ((\frac{π}{4} + \frac{1}{2} \cos^{- 1} ((\frac{2}{3})))) + \tan ((\frac{1}{2} \sin^{- 1} ((\frac{2}{3})))),$ then the number of solutions of the equation $\sin^{- 1} (k x - 1) = \sin^{- 1} x - \cos^{- 1} x$ is: [2026]

(1)

$Let θ = \frac{1}{2} \sin^{- 1} (\frac{2}{3}), then \frac{1}{2} \cos^{- 1} (\frac{1}{3}) = (\frac{π}{4} - θ)$

$k = \tan θ + \cot θ = \frac{1}{\sin θ \cos θ} = \frac{2}{\sin 2 θ}$

$k = \frac{2}{2 / 3} = 3$

$\sin^{- 1} (3 x - 1) = \sin^{- 1} x - \cos^{- 1} x$

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

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

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

$No. of solution = 1$

Q 22 :

If $\cos (\sin^{- 1} \frac{2}{5} + \cos^{- 1} x) = 0,$ then $x$ is equal to:

$\frac{1}{5}$
$\frac{2}{5}$
$0$
$1$

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(2)

Ans. $\frac{2}{5}$

Explanation:

We have,

$\cos (\sin^{- 1} \frac{2}{5} + \cos^{- 1} x) = 0$

$\Rightarrow$ $\sin^{- 1} \frac{2}{5} + \cos^{- 1} x = \cos^{- 1} 0$

$\Rightarrow$ $\sin^{- 1} \frac{2}{5} + \cos^{- 1} x = \frac{π}{2}$

$\Rightarrow$ $\cos^{- 1} x = \frac{π}{2} - \sin^{- 1} \frac{2}{5}$

$\Rightarrow$ $\cos^{- 1} x = \cos^{- 1} \frac{2}{5}$ $(∵ \cos^{- 1} x + \sin^{- 1} x = \frac{π}{2})$

$∴$ $x = \frac{2}{5}$

Q 23 :

Which of the following corresponds to the principal value branch of $\tan^{- 1} x$ ?

$(- \frac{π}{2}, \frac{π}{2})$
$[- \frac{π}{2}, \frac{π}{2}]$
$(- \frac{π}{2}, \frac{π}{2}) - {0}$
$(0, π)$

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(1)

Ans. $(- \frac{π}{2}, \frac{π}{2})$

Q 24 :

The principal value of the expression $\cos^{- 1} [\cos (- 680 °)]$ is:

$\frac{2 π}{9}$
$\frac{- 2 π}{9}$
$\frac{34 π}{9}$
$\frac{π}{9}$

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(1)

Ans. $\frac{2 π}{9}$

Explanation:

$\cos^{- 1} [\cos (- 680 °)] = \cos^{- 1} [\cos (720 ° - 40 °)]$

$= \cos^{- 1} [\cos (- 40 °)]$

$= \cos^{- 1} [\cos (40 °)]$

$= 40 ° = \frac{2 π}{9}$

Q 25 :

The domain of $\sin^{- 1} 2 x$ is:

$[0, 1]$
$[- 1, 1]$
$[- \frac{1}{2}, \frac{1}{2}]$
$[- 2, 2]$

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(3)

Ans. $[- \frac{1}{2}, \frac{1}{2}]$

Explanation:

Let $\sin^{- 1} 2 x = θ$

So, $2 x = \sin θ$

Now, $- 1 \leq \sin θ \leq 1,$ i.e., $- 1 \leq 2 x \leq 1,$ which gives $- \frac{1}{2} \leq x \leq \frac{1}{2}$

Q 26 :

The domain of $y = \cos^{- 1} (x^{2} - 4)$ is:

$[3, 5]$
$[0, π]$
$[- \sqrt{5}, - \sqrt{3}] \cap [- \sqrt{5}, \sqrt{3}]$
$[- \sqrt{5}, - \sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$

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(4)

Ans. $[- \sqrt{5}, - \sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$

Explanation:

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

$\Rightarrow$ $\cos y = x^{2} - 4$

i.e., $- 1 \leq x^{2} - 4 \leq 1$ (since $- 1 \leq \cos y \leq 1$ )

$\Rightarrow 3 \leq x^{2} \leq 5$

$\Rightarrow \sqrt{3} \leq | x | \leq \sqrt{5}$

$\Rightarrow x \in [- \sqrt{5}, - \sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$

Q 27 :

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

$\frac{π}{2}$
$\frac{π}{6}$
$\frac{π}{3}$
$π$

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(2)

Ans. $\frac{π}{6}$

Explanation:

$\tan^{- 1} (\frac{1}{\sqrt{3}}) = \tan^{- 1} (\tan \frac{π}{6}) = \frac{π}{6}$

Q 28 :

The inverse of cosine function is defined in the intervals:

$[- π, 0]$
$[\frac{- π}{2}, 0]$
$[0, \frac{π}{2}]$
$[\frac{π}{2}, π]$

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(1)

Ans. $[- π, 0]$

Explanation:

Cosine function with respect to any interval $[- π, 0]$ , $[0, π]$ , $[π, 2 π]$ etc., is bijective with range $[- 1, 1]$ .

Q 29 :

$\sin (\frac{π}{3} - \sin^{- 1} (- \frac{1}{2}))$ is equal to:

$\frac{1}{\sqrt{2}}$
$\frac{1}{3}$
$\frac{\sqrt{3}}{2}$
$1$

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(4)

Ans. 1

Explanation:

$\sin [\frac{π}{3} - \sin^{- 1} (- \frac{1}{2})]$

$= \sin [\frac{π}{3} + \frac{π}{6}]$ $[∵ \sin^{- 1} (- \frac{1}{2}) = - \frac{π}{6}]$

$= \sin \frac{π}{2} = 1$

Q 30 :

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

[1, 2]
[−1, 1]
[0, 1]
None of these

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(1)

Ans: [1, 2]

Explanation:

We know that $\sin^{- 1} x$ is defined for

$x \in [- 1, 1]$

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

$\Rightarrow$ $0 \leq \sqrt{x - 1} \leq 1$

$[∵ \sqrt{x - 1} \geq 0 and - 1 \leq \sqrt{x - 1} \leq 1]$

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

$\Rightarrow 1 \leq x \leq 2$

$∴$ $x \in [1, 2]$

Q 31 :

The value of $\cot (\sin^{- 1} x)$ is:

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

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(4)

Ans. $\frac{\sqrt{1 - x^{2}}}{x}$

Explanation:

Let $\sin^{- 1} x = θ$ ,

then $\sin θ = x$

$\Rightarrow$ $c o s e c θ = \frac{1}{x}$

$\Rightarrow$ $c o s e c^{2} θ = \frac{1}{x^{2}}$

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

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

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

Q 32 :

The value of $\tan {\cos^{- 1} (\frac{1}{5 \sqrt{2}}) - \sin^{- 1} (\frac{4}{\sqrt{17}})}$ is:

$\frac{\sqrt{29}}{3}$
$\frac{29}{3}$
$\frac{\sqrt{3}}{29}$
$\frac{3}{29}$

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(4)

Ans. $\frac{3}{29}$

Explanation:

$\tan {\cos^{- 1} (\frac{1}{5 \sqrt{2}}) - \sin^{- 1} (\frac{4}{\sqrt{17}})}$

Let $A = \cos^{- 1} (\frac{1}{5 \sqrt{2}})$

$∴$ $\cos A = \frac{1}{5 \sqrt{2}}$

$\Rightarrow$ $\cos^{2} A = \frac{1}{50}$

$\Rightarrow$ $\sec^{2} A = 50$

$\Rightarrow$ $\tan^{2} A = 50 - 1 = 49$

and $B = \sin^{- 1} (\frac{4}{\sqrt{17}})$

$\sin B = \frac{4}{\sqrt{17}}$

$\sin^{2} B = \frac{16}{17}$

$c o s e c^{2} B = \frac{17}{16}$

$\cot^{2} B = \frac{17}{16} - 1 = \frac{1}{16}$

$∴$ $\tan A = 7 and \tan B = 4$

Now, $\tan {\cos^{- 1} (\frac{1}{5 \sqrt{2}}) - \sin^{- 1} (\frac{4}{\sqrt{17}})}$

$= \tan (A - B)$

$= \frac{\tan A - \tan B}{1 + \tan A \tan B}$

$= \frac{7 - 4}{1 + 7 \times 4} = \frac{3}{29}$

Q 33 :

The value of $\tan^{2} (\sec^{- 1} 2) + \cot^{2} (c o s e c^{- 1} 3)$ is:

5
11
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15

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(2)

Ans. 11

Explanation:

$\tan^{2} (\sec^{- 1} 2) + \cot^{2} (c o s e c^{- 1} 3)$

$= \sec^{2} (\sec^{- 1} 2) - 1 + c o s e c^{2} (c o s e c^{- 1} 3) - 1$

$= 2^{2} + 3^{2} - 2$

$= 11$

Q 34 :

If $\sin^{- 1} x + \sin^{- 1} y = \frac{2 π}{3},$ then the value of $\cos^{- 1} x + \cos^{- 1} y$ will be:

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

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(2)

Ans. $\frac{π}{3}$

Explanation:

$\sin^{- 1} x + \sin^{- 1} y = \frac{2 π}{3}$

$\Rightarrow (\frac{π}{2} - \cos^{- 1} x) + (\frac{π}{2} - \cos^{- 1} y) = \frac{2 π}{3}$ $(∵ \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2})$

$\Rightarrow$ $π - \frac{2 π}{3} = \cos^{- 1} x + \cos^{- 1} y$

$∴$ $\cos^{- 1} x + \cos^{- 1} y = \frac{π}{3}$

Q 35 :

$\cot^{- 1} (\frac{a b + 1}{a - b}) + \cot^{- 1} (\frac{b c + 1}{b - c}) + \cot^{- 1} (\frac{c a + 1}{c - a}) =$

$0$
$1$
$\frac{π}{4}$
$- 1$

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(1)

Ans. 0

Explanation:

$\cot^{- 1} (\frac{a b + 1}{a - b}) + \cot^{- 1} (\frac{b c + 1}{b - c}) + \cot^{- 1} (\frac{c a + 1}{c - a})$

$= \tan^{- 1} (\frac{a - b}{a b + 1}) + \tan^{- 1} (\frac{b - c}{1 + b c}) + \tan^{- 1} (\frac{c - a}{1 + c a})$

$= \tan^{- 1} a - \tan^{- 1} b + \tan^{- 1} b - \tan^{- 1} c + \tan^{- 1} c - \tan^{- 1} a = 0$

Q 36 :

If $\tan^{- 1} x + \tan^{- 1} y + \tan^{- 1} z = π,$ then $x + y + z =$

$x y z$
$1$
$0$
$\frac{1}{x y z}$

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(1)

Ans. $x y z$

Explanation:

Given,

$\tan^{- 1} x + \tan^{- 1} y + \tan^{- 1} z = π$

$\Rightarrow$ $\tan^{- 1} x + \tan^{- 1} y = π - \tan^{- 1} z$

$\Rightarrow$ $\tan^{- 1} (\frac{x + y}{1 - x y}) = π - \tan^{- 1} z$

$\Rightarrow$ $\frac{x + y}{1 - x y} = \tan (π - \tan^{- 1} z)$

$\Rightarrow$ $\frac{x + y}{1 - x y} = - \tan (\tan^{- 1} z) = - z$

$\Rightarrow$ $x + y = - z + x y z$

$\Rightarrow$ $x + y + z = x y z$

Q 37 :

If $\tan^{- 1} (\frac{x - 1}{x + 2}) + \tan^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{4},$ then $x =$

$\frac{1}{\sqrt{2}}$
$- \frac{1}{\sqrt{2}}$
$\pm \sqrt{\frac{5}{2}}$
$\pm \frac{1}{2}$

1

2

3

4

(3)

Ans. $\pm \sqrt{\frac{5}{2}}$

Explanation:

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

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

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

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

$\Rightarrow$ $2 x^{2} + 4 x = 4 x + 5$

$\Rightarrow$ $2 x^{2} = 5$

$\Rightarrow$ $x^{2} = \frac{5}{2}$

$∴$ $x = \pm \sqrt{\frac{5}{2}}$

Q 38 :

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

$\sin^{- 1} x + \sin^{- 1} \sqrt{x}$
$\sin^{- 1} x - \sin^{- 1} \sqrt{x}$
$\sin^{- 1} \sqrt{x} - \sin^{- 1} x$
None of these

1

2

3

4

(2)

Ans. $s i n^{- 1} x - s i n^{- 1} \sqrt{x}$

Explanation:

We know that

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

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

$= \sin^{- 1} x - \sin^{- 1} \sqrt{x}$

Q 39 :

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

$\frac{π}{3}$
$\frac{2 π}{3}$
$- \frac{π}{3}$
$\frac{π}{6}$

1

2

3

4

(1)

Ans. $\frac{π}{3}$

Explanation:

$\tan^{- 1} [2 \sin {2 \cos^{- 1} (\frac{\sqrt{3}}{2})}]$

$= \tan^{- 1} [2 \sin (2 \times \frac{π}{6})]$

$= \tan^{- 1} [2 \sin \frac{π}{3}]$

$= \tan^{- 1} [2 \times \frac{\sqrt{3}}{2}]$

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

$= \frac{π}{3}$

Q 40 :

If $\sin^{- 1} x + \sin^{- 1} y = \frac{2 π}{3},$ then the value of $\cos^{- 1} x + \cos^{- 1} y$ is:

$\frac{π}{2}$
$\frac{π}{3}$
$\frac{2 π}{3}$
$\frac{π}{6}$

1

2

3

4

(2)

Ans. $\frac{π}{3}$

Explanation:

$\sin^{- 1} x + \sin^{- 1} y = \frac{2 π}{3}$

$\Rightarrow$ $(\frac{π}{2} - \cos^{- 1} x) + (\frac{π}{2} - \cos^{- 1} y) = \frac{2 π}{3}$ $[∵ \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}]$

$\Rightarrow$ $\cos^{- 1} x + \cos^{- 1} y = π - \frac{2 π}{3} = \frac{π}{3}$

Topic Question Set

Q 11 :

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 $α x^{2} + β x + \sin^{- 1} (x^{2} - 6 x + 10) + \cos^{- 1} (x^{2} - 6 x + 10) = 0$ and $α - β = b - a,$ then $α$ is equal to [2023]

Q 12 :

If the domain of the function $f (x) = \sec^{- 1} (\frac{2 x}{5 x + 3})$ is $[α, β) \cup (γ, δ],$ then $| 3 α + 10 (β + γ) + 21 δ |$ is equal to ______ . [2023]

Q 13 :

If $S = {x \in ℝ : \sin^{- 1} (\frac{x + 1}{\sqrt{x^{2} + 2 x + 2}}) - \sin^{- 1} (\frac{x}{\sqrt{x^{2} + 1}}) = \frac{π}{4}},$ then $\sum_{x \in S} (\sin ((x^{2} + x + 5) \frac{π}{2}) - \cos ((x^{2} + x + 5) π))$ is equal to _____ . [2023]

Q 14 :

For $x \in (- 1, 1],$ the number of solutions of the equation $\sin^{- 1} x = 2 \tan^{- 1} x$ is equal to ________ . [2023]

Q 15 :

If the sum of all the solutions of $\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{π}{3}, - 1 < x < 1, x \neq 0,$ is $α - \frac{4}{\sqrt{3}},$ then $α$ is equal to _______ . [2023]

Q 16 :

If the domain of the function $f (x)$ $= \cos^{- 1} (\frac{2 x - 5}{11 - 3 x}) + \sin^{- 1} (2 x^{2} - 3 x + 1)$ is the interval $[α, β]$ then $α + 2 β$ is equal to: [2026]

Q 17 :

The number of solutions of $\tan^{- 1} 4 x + \tan^{- 1} 6 x = \frac{π}{6},$ where $- \frac{1}{2 \sqrt{6}} < x < \frac{1}{2 \sqrt{6}},$ is equal to [2026]

Q 18 :

Considering the principal values of inverse trigonometric functions, the value of the expression $\tan (2 \sin^{- 1} (\frac{2}{\sqrt{13}}) - 2 \cos^{- 1} (\frac{3}{\sqrt{10}}))$ is equal to: [2026]

Q 19 :

If the domain of the function $f (x)$ $= \sin^{- 1} (\frac{1}{x^{2} - 2 x - 2})$ , is $(- \infty, α] \cup [β, γ] \cup [δ, \infty),$ then $α + β + γ + δ$ is equal to [2026]

Q 20 :

Let the maximum value of ${(\sin^{- 1} x)}^{2} + {(\cos^{- 1} x)}^{2}$ for $x \in [- \frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}]$ be $\frac{m}{n} π^{2}$ , where $\gcd (m, n) = 1$ . Then $m + n$ is equal to_____ [2026]

Q 21 :

If $k = \tan ((\frac{π}{4} + \frac{1}{2} \cos^{- 1} ((\frac{2}{3})))) + \tan ((\frac{1}{2} \sin^{- 1} ((\frac{2}{3})))),$ then the number of solutions of the equation $\sin^{- 1} (k x - 1) = \sin^{- 1} x - \cos^{- 1} x$ is: [2026]

Q 22 :

If $\cos (\sin^{- 1} \frac{2}{5} + \cos^{- 1} x) = 0,$ then $x$ is equal to:

Q 23 :

Which of the following corresponds to the principal value branch of $\tan^{- 1} x$ ?

Q 24 :

The principal value of the expression $\cos^{- 1} [\cos (- 680 °)]$ is:

Q 25 :

The domain of $\sin^{- 1} 2 x$ is:

Q 26 :

The domain of $y = \cos^{- 1} (x^{2} - 4)$ is:

Q 27 :

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

Q 28 :

The inverse of cosine function is defined in the intervals:

Q 29 :

$\sin (\frac{π}{3} - \sin^{- 1} (- \frac{1}{2}))$ is equal to:

Q 30 :

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

Q 31 :

The value of $\cot (\sin^{- 1} x)$ is:

Q 32 :

The value of $\tan {\cos^{- 1} (\frac{1}{5 \sqrt{2}}) - \sin^{- 1} (\frac{4}{\sqrt{17}})}$ is:

Q 33 :

The value of $\tan^{2} (\sec^{- 1} 2) + \cot^{2} (c o s e c^{- 1} 3)$ is:

Q 34 :

If $\sin^{- 1} x + \sin^{- 1} y = \frac{2 π}{3},$ then the value of $\cos^{- 1} x + \cos^{- 1} y$ will be:

Q 35 :

$\cot^{- 1} (\frac{a b + 1}{a - b}) + \cot^{- 1} (\frac{b c + 1}{b - c}) + \cot^{- 1} (\frac{c a + 1}{c - a}) =$

Q 36 :

If $\tan^{- 1} x + \tan^{- 1} y + \tan^{- 1} z = π,$ then $x + y + z =$

Q 37 :

If $\tan^{- 1} (\frac{x - 1}{x + 2}) + \tan^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{4},$ then $x =$

Q 38 :

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

Q 39 :

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

Q 40 :

If $\sin^{- 1} x + \sin^{- 1} y = \frac{2 π}{3},$ then the value of $\cos^{- 1} x + \cos^{- 1} y$ is:

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Topic Question Set

Q 11 : Let (a,b)⊂(0,2π) be the largest interval for which sin-1(sinθ)-cos-1(sinθ)>0, θ∈(0,2π), holds. If αx2+βx+sin-1(x2-6x+10)+cos-1(x2-6x+10)=0 and α-β=b-a, then α is equal to [2023]

Q 12 : If the domain of the function f(x)=sec-1(2x5x+3) is [α,β)∪(γ,δ], then |3α+10(β+γ)+21δ| is equal to ______ . [2023]

Q 13 : If S={x∈ℝ: sin-1(x+1x2+2x+2)-sin-1(xx2+1)=π4}, then ∑x∈S(sin((x2+x+5)π2)-cos((x2+x+5)π)) is equal to _____ . [2023]

Q 14 : For x∈(-1,1], the number of solutions of the equation sin-1x=2tan-1x is equal to ________ . [2023]

Q 15 : If the sum of all the solutions of tan-1(2x1-x2)+cot-1(1-x22x)=π3,-1<x<1,x≠0, is α-43, then α is equal to _______ . [2023]

Q 16 : If the domain of the function f(x)=cos-1(2x-511-3x)+sin-1(2x2-3x+1) is the interval [α,β] then α+2β is equal to: [2026]

Q 17 : The number of solutions of tan-14x+tan-16x=π6, where -126<x<126, is equal to [2026]

Q 18 : Considering the principal values of inverse trigonometric functions, the value of the expression tan(2sin-1(213)-2cos-1(310)) is equal to: [2026]

Q 19 : If the domain of the function f(x)=sin-1(1x2-2x-2), is (-∞,α]∪[β,γ]∪[δ,∞), then α+β+γ+δ is equal to [2026]

Q 20 : Let the maximum value of (sin-1x)2+(cos-1x)2 for x∈[-32,12] be mnπ2, where gcd(m,n)=1. Then m+n is equal to_____ [2026]

Q 21 : If k=tan(π4+12cos-1(23))+tan(12sin-1(23)), then the number of solutions of the equation sin-1(kx-1)=sin-1x-cos-1x is: [2026]

Q 22 : If cos(sin-125+cos-1x)=0, then x is equal to:

Q 23 : Which of the following corresponds to the principal value branch of tan-1x?

Q 24 : The principal value of the expression cos-1[cos(-680°)] is:

Q 25 : The domain of sin-12x is:

Q 26 : The domain of y=cos-1(x2-4) is:

Q 27 : The principal value of tan-1(13) is:

Q 28 : The inverse of cosine function is defined in the intervals:

Q 29 : sin(π3-sin-1(-12)) is equal to:

Q 30 : The domain of the function defined by f(x)=sin-1(x−1) is:

Q 31 : The value of cot(sin-1x) is:

Q 32 : The value of tan{cos-1(152)-sin-1(417)} is:

Q 33 : The value of tan2(sec-12)+cot2(cosec-13) is:

Q 34 : If sin-1x+sin-1y=2π3, then the value of cos-1x+cos-1y will be:

Q 35 : cot-1(ab+1a-b)+cot-1(bc+1b-c)+cot-1(ca+1c-a)=

Q 36 : If tan-1x+tan-1y+tan-1z=π, then x+y+z=

Q 37 : If tan-1(x-1x+2)+tan-1(x+1x+2)=π4, then x=

Q 38 : sin-1(x1-x-x1-x2)=…………

Q 39 : The value of tan-1[2sin{2cos-1(32)}] is:

Q 40 : If sin-1x+sin-1y=2π3, then the value of cos-1x+cos-1y is:

Want Us to Email you About Special Offers & Updates?

Q 11 :

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 $α x^{2} + β x + \sin^{- 1} (x^{2} - 6 x + 10) + \cos^{- 1} (x^{2} - 6 x + 10) = 0$ and $α - β = b - a,$ then $α$ is equal to [2023]

Q 12 :

If the domain of the function $f (x) = \sec^{- 1} (\frac{2 x}{5 x + 3})$ is $[α, β) \cup (γ, δ],$ then $| 3 α + 10 (β + γ) + 21 δ |$ is equal to ______ . [2023]

Q 13 :

If $S = {x \in ℝ : \sin^{- 1} (\frac{x + 1}{\sqrt{x^{2} + 2 x + 2}}) - \sin^{- 1} (\frac{x}{\sqrt{x^{2} + 1}}) = \frac{π}{4}},$ then $\sum_{x \in S} (\sin ((x^{2} + x + 5) \frac{π}{2}) - \cos ((x^{2} + x + 5) π))$ is equal to _____ . [2023]

Q 14 :

For $x \in (- 1, 1],$ the number of solutions of the equation $\sin^{- 1} x = 2 \tan^{- 1} x$ is equal to ________ . [2023]

Q 15 :

If the sum of all the solutions of $\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{π}{3}, - 1 < x < 1, x \neq 0,$ is $α - \frac{4}{\sqrt{3}},$ then $α$ is equal to _______ . [2023]

Q 16 :

If the domain of the function $f (x)$ $= \cos^{- 1} (\frac{2 x - 5}{11 - 3 x}) + \sin^{- 1} (2 x^{2} - 3 x + 1)$ is the interval $[α, β]$ then $α + 2 β$ is equal to: [2026]

Q 17 :

The number of solutions of $\tan^{- 1} 4 x + \tan^{- 1} 6 x = \frac{π}{6},$ where $- \frac{1}{2 \sqrt{6}} < x < \frac{1}{2 \sqrt{6}},$ is equal to [2026]

Q 18 :

Considering the principal values of inverse trigonometric functions, the value of the expression $\tan (2 \sin^{- 1} (\frac{2}{\sqrt{13}}) - 2 \cos^{- 1} (\frac{3}{\sqrt{10}}))$ is equal to: [2026]

Q 19 :

If the domain of the function $f (x)$ $= \sin^{- 1} (\frac{1}{x^{2} - 2 x - 2})$ , is $(- \infty, α] \cup [β, γ] \cup [δ, \infty),$ then $α + β + γ + δ$ is equal to [2026]

Q 20 :

Let the maximum value of ${(\sin^{- 1} x)}^{2} + {(\cos^{- 1} x)}^{2}$ for $x \in [- \frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}]$ be $\frac{m}{n} π^{2}$ , where $\gcd (m, n) = 1$ . Then $m + n$ is equal to_____ [2026]

Q 21 :

If $k = \tan ((\frac{π}{4} + \frac{1}{2} \cos^{- 1} ((\frac{2}{3})))) + \tan ((\frac{1}{2} \sin^{- 1} ((\frac{2}{3})))),$ then the number of solutions of the equation $\sin^{- 1} (k x - 1) = \sin^{- 1} x - \cos^{- 1} x$ is: [2026]

Q 22 :

If $\cos (\sin^{- 1} \frac{2}{5} + \cos^{- 1} x) = 0,$ then $x$ is equal to:

Q 23 :

Which of the following corresponds to the principal value branch of $\tan^{- 1} x$ ?

Q 24 :

The principal value of the expression $\cos^{- 1} [\cos (- 680 °)]$ is:

Q 25 :

The domain of $\sin^{- 1} 2 x$ is:

Q 26 :

The domain of $y = \cos^{- 1} (x^{2} - 4)$ is:

Q 27 :

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

Q 28 :

The inverse of cosine function is defined in the intervals:

Q 29 :

$\sin (\frac{π}{3} - \sin^{- 1} (- \frac{1}{2}))$ is equal to:

Q 30 :

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

Q 31 :

The value of $\cot (\sin^{- 1} x)$ is:

Q 32 :

The value of $\tan {\cos^{- 1} (\frac{1}{5 \sqrt{2}}) - \sin^{- 1} (\frac{4}{\sqrt{17}})}$ is:

Q 33 :

The value of $\tan^{2} (\sec^{- 1} 2) + \cot^{2} (c o s e c^{- 1} 3)$ is:

Q 34 :

If $\sin^{- 1} x + \sin^{- 1} y = \frac{2 π}{3},$ then the value of $\cos^{- 1} x + \cos^{- 1} y$ will be:

Q 35 :

$\cot^{- 1} (\frac{a b + 1}{a - b}) + \cot^{- 1} (\frac{b c + 1}{b - c}) + \cot^{- 1} (\frac{c a + 1}{c - a}) =$

Q 36 :

If $\tan^{- 1} x + \tan^{- 1} y + \tan^{- 1} z = π,$ then $x + y + z =$

Q 37 :

If $\tan^{- 1} (\frac{x - 1}{x + 2}) + \tan^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{4},$ then $x =$

Q 38 :

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

Q 39 :

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

Q 40 :

If $\sin^{- 1} x + \sin^{- 1} y = \frac{2 π}{3},$ then the value of $\cos^{- 1} x + \cos^{- 1} y$ is: