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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Mixed difficulty

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451 Hard
451 of 605
Application +4 | -1
If the solution set of system of equations 3sin⁡−1x=−π3 \sin ^{-1} x=-\pi −sin⁡−1(3x−4x3)-\sin ^{-1}\left(3 x-4 x^3\right) is [a,b][a, b] and that of 3sin⁡−1x=π−sin⁡−13 \sin ^{-1} x=\pi-\sin ^{-1} (3x−4x3)\left(3 x-4 x^3\right) is [c,d][\mathrm{c}, \mathrm{d}] then evaluate ∣a∣+∣b∣+∣c∣+∣d∣|a|+|b|+|c|+|d|
View answer and explanation
Correct option: A
3
Option A (3) is correct because the solution sets are [-1, -1/2] and [1/2, 1], so |a| + |b| + |c| + |d| = 3.
452 Medium
452 of 605
Application +4 | -1
Let a∈(−π2,π2)a \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right) such that tan⁡−1(tan⁡α3+2tan⁡2α)+tan⁡−1(2tan⁡α3)=π12\tan ^{-1}\left(\frac{\tan \alpha}{3+2 \tan ^2 \alpha}\right)+\tan ^{-1}\left(\frac{2 \tan \alpha}{3}\right)=\frac{\pi}{12}, then α\alpha equals:
View answer and explanation
Correct option: D
π12\frac{\pi}{12}
Option D (π12\frac{\pi}{12}) is correct because by simplifying the given equation using trigonometric identities, we find that α\alpha equals π12\frac{\pi}{12}.
453 Easy
453 of 605
Application +4 | -1
tan⁡−1(sin⁡1−1cos⁡1)\tan ^{-1}\left(\frac{\sin 1-1}{\cos 1}\right) equals
View answer and explanation
Correct option: D
12−π4\frac{1}{2}-\frac{\pi}{4}
Option D (12−π4\frac{1}{2}-\frac{\pi}{4}) is correct because using the half-angle formula and simplifying leads to this result.
454 Easy
454 of 605
Understanding +4 | -1
The value of sin⁡(tan⁡−1x+cot⁡−1x)sin⁡(sin⁡−1x+cos⁡−1x)\frac{\sin \left(\tan ^{-1} x+\cot ^{-1} x\right)}{\sin \left(\sin ^{-1} x+\cos ^{-1} x\right)} is
View answer and explanation
Correct option: A
1
The correct answer is A (1) because tan⁡−1x+cot⁡−1x=π2\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2} and sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}, so the expression simplifies to sin⁡(π2)sin⁡(π2)=1\frac{\sin(\frac{\pi}{2})}{\sin(\frac{\pi}{2})} = 1.
455 Easy
455 of 605
Application +4 | -1
The principal value of sin⁡−1(sin⁡5π3)\sin ^{-1}\left(\sin \frac{5 \pi}{3}\right) is
View answer and explanation
Correct option: C
−π3-\frac{\pi}{3}
Option C (−π3-\frac{{\pi}}{{3}}) is correct because it is the principal value of the inverse sine function for the given value.
456 Easy
456 of 605
Application +4 | -1
The value of tan⁡[sin⁡−1(7/50)+cos⁡−1(1/50)+cot⁡−1(1/7)]\tan \left[\sin ^{-1}(7 / \sqrt{50})+\cos ^{-1}(1 / \sqrt{50})+\cot ^{-1}(1 / 7)\right] is equal to
View answer and explanation
Correct option: C
161/73161 / 73
Option C (161/73161 / 73) is the correct answer because it is the simplified result of applying trigonometric identities and the tangent addition formula.
457 Medium
457 of 605
Analyzing +4 | -1
The number of real solution of cot⁡−1x(x+4)+cos⁡−1x2+4x+1=π2\cot ^{-1} \sqrt{x(x+4)}+\cos ^{-1} \sqrt{x^2+4 x+1}=\frac{\pi}{2} is equal to
View answer and explanation
Correct option: C
2
Option C (2) is correct because the equation is satisfied only when x2+4x=0x^2 + 4x = 0, leading to two real solutions: 0 and -4.
458 Easy
458 of 605
Application +4 | -1
The value of cos⁡(2cos⁡−1x+sin⁡−1x)\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right) at x=16x=\frac{1}{6} :
View answer and explanation
Correct option: B
−356-\frac{\sqrt{35}}{6}
Option B (−356-\frac{\sqrt{35}}{6}) is correct because the expression simplifies to −1−x2-\sqrt{1-x^2} and substituting x=16x=\frac{1}{6} gives the result.
459 Easy
459 of 605
Application +4 | -1
If x>−13x > - \frac{1}{{\sqrt 3 }}, then tan⁡−1(3x−x31−3x2){\tan ^{ - 1}}\left( {\frac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right) equals
View answer and explanation
Correct option: C
π+3tan⁡−1x\pi + 3{\tan ^{ - 1}}x
Option C (π+3tan⁡−1x\pi + 3{\tan ^{ - 1}}x) is correct because it accounts for the range of the arctangent function when x>−13x > - \frac{1}{{\sqrt 3 }} and applies the triple angle formula for tangent.
460 Medium
460 of 605
Application +4 | -1
If −12≤x≤12- \frac{1}{2} \le x \le \frac{1}{2}, then cos⁡−1(4x3−3x){\cos ^{ - 1}}\left( {4{x^3} - 3x} \right) equals
View answer and explanation
Correct option: B
2π−3cos⁡−1x2\pi - 3{\cos ^{ - 1}}x
Option D (None of these) is correct because the correct expression is 3cos⁡−1x3{\cos ^{ - 1}}x when 0≤x≤10 \leq x \leq 1, and 2π−3cos⁡−1x2\pi - 3{\cos ^{ - 1}}x when −1≤x≤−1/2-1 \leq x \leq -1/2. Since the given range is −1/2≤x≤1/2-1/2 \le x \le 1/2, the expression is 3cos⁡−1x3\cos^{-1}x only when 1/2≤x≤11/2 \le x \le 1.
461 Hard
461 of 605
Application +4 | -1
tan⁡[12cos⁡−1(53)]\tan \left[\frac{1}{2} \cos ^{-1}\left(\frac{\sqrt{5}}{3}\right)\right] is equal to
View answer and explanation
Correct option: A
3−52\frac{3-\sqrt{5}}{2}
Option A (3−52\frac{3-\sqrt{5}}{2}) is correct because using the half-angle formula for tangent and simplifying the expression leads to the correct answer.
462 Medium
462 of 605
Application +4 | -1
The xx satisfying sin⁡−1x+sin⁡−1(1−x)=cos⁡−1x\sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x are
View answer and explanation
Correct option: C
0,120, \frac{1}{2}
Option C (0,120, \frac{{1}}{{2}}) is correct because these values satisfy the given equation when substituted and verified using inverse trigonometric properties.
463 Medium
463 of 605
Application +4 | -1
If x<0x < 0, then tan⁡−1x+tan⁡−1(1/x)\tan ^{-1} x+\tan ^{-1}(1 / x) equals
View answer and explanation
Correct option: C
−π/2-\pi / 2
The correct answer is C (−π/2-\pi / 2) because when x < 0, the identity tan⁡−1x+tan⁡−1(1/x)=−π/2\tan^{-1} x + \tan^{-1}(1/x) = -\pi/2 holds.
464 Medium
464 of 605
Application +4 | -1
If 12sin⁡−1[3sin⁡2θ5+4cos⁡2θ]=tan⁡−1x\frac{1}{2} \sin ^{-1}\left[\frac{3 \sin 2 \theta}{5+4 \cos 2 \theta}\right]=\tan ^{-1} x, then x=x=
View answer and explanation
Correct option: C
(1/3)tan⁡θ(1 / 3) \tan \theta
Option C ((1/3)tan θ) is correct because by simplifying the expression inside the inverse sine function, we can find the value of x.
465 Easy
465 of 605
Application +4 | -1
The value of tan⁡−11+tan⁡−12+tan⁡−13{\tan ^{ - 1}}1 + {\tan ^{ - 1}}2 + {\tan ^{ - 1}}3, is
View answer and explanation
Correct option: C
π\pi
Option C (π\pi) is correct because by applying the formula tan⁡−1x+tan⁡−1y=π+tan⁡−1(x+y1−xy){\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \pi + {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right) when xy>1xy > 1, the expression simplifies to π\pi.

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466 Medium
466 of 605
Application +4 | -1
If sin⁡−1(sin⁡x)+cos⁡−1(cos⁡y)+sec⁡−1(sec⁡z)=5π2\sin ^{-1}(\sin x)+\cos ^{-1}(\cos y)+\sec ^{-1}(\sec z)=\frac{5 \pi}{2} then the value of sin⁡x+cos⁡y+sec⁡z\sin x+\cos y+\sec z is
View answer and explanation
Correct option: B
-1
Option B (-1) is correct because the given equation is only possible if sin x = 1, cos y = -1, and sec z = -1, leading to sin x + cos y + sec z = -1.
467 Easy
467 of 605
Application +4 | -1
If 3sin⁡−12x1+x2−4cos⁡−11+x1+x2+2tan⁡−12x1−x2=π3,3{\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} - 4{\cos ^{ - 1}}\frac{{1 + x}}{{1 + {x^2}}} + 2{\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} = \frac{\pi }{3}, then value of xx is
View answer and explanation
Correct option: B
13\frac{1}{{\sqrt 3 }}
Option B (13\frac{1}{{\sqrt 3 }}) is correct because substituting x=tan⁡θx = \tan \theta and simplifying the equation leads to x=13x = \frac{1}{{\sqrt 3 }}.
468 Medium
468 of 605
Application +4 | -1
If sin⁡−12a1+a2+sin⁡−12b1+b2=2tan⁡−1x,\sin ^{-1} \frac{2 a}{1+a^2}+\sin ^{-1} \frac{2 b}{1+b^2}=2 \tan ^{-1} x, then xx is equal to
View answer and explanation
Correct option: D
a+b1−ab\frac{a+b}{1-a b}
The correct answer is B (x+2y+4=0x+2 y+4=0) because it represents a common tangent to both the parabola and the ellipse.
469 Easy
469 of 605
Application +4 | -1
tan⁡−1(3x−x31−3x2)−tan⁡−1(2x1−x2){\tan ^{ - 1}}\left( {\frac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right) - {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right) is equal to
View answer and explanation
Correct option: C
tan⁡−1x{\tan ^{ - 1}}x
Option C (tan⁡−1x{\tan ^{ - 1}}x) is correct because by substituting x=tan⁡θx = \tan{\theta}, the expression simplifies to 3tan⁡−1x−2tan⁡−1x3{\tan ^{ - 1}}x - 2{\tan ^{ - 1}}x, which equals tan⁡−1x{\tan ^{ - 1}}x.
470 Medium
470 of 605
Application +4 | -1
The number of solutions of equation 2(sin⁡−1x)2−5sin⁡−1x+2=02\left(\sin ^{-1} x\right)^2-5 \sin ^{-1} x+2=0 is
View answer and explanation
Correct option: B
1
Option B (1) is correct because only one solution for xx exists within the domain of sin⁡−1x\sin^{-1} x, which is [−1,1][-1, 1].
471 Easy
471 of 605
Application +4 | -1
The value of tan {cos⁡−1(−27)−π2}\left\{ {{{\cos }^{ - 1}}\left( { - \frac{2}{7}} \right) - \frac{\pi }{2}} \right\} is
View answer and explanation
Correct option: A
235\frac{2}{{3\sqrt 5 }}
Option A (235\frac{2}{{3\sqrt 5 }}) is correct because it is the simplified value of the given trigonometric expression after applying trigonometric identities.
472 Medium
472 of 605
Application +4 | -1
If we consider only the principle value of the inverse trigonometric functions, then the value of tan⁡(cos⁡−1152−sin⁡−1417)\tan \left( {{{\cos }^{ - 1}}\frac{1}{{5\sqrt 2 }} - {{\sin }^{ - 1}}\frac{4}{{\sqrt {17} }}} \right) is
View answer and explanation
Correct option: D
329\frac{3}{{29}}
Option D (329\frac{3}{{29}}) is correct because the expression simplifies using trigonometric identities and the tangent subtraction formula.
473 Easy
473 of 605
Remembering +4 | -1
The principal value of sin⁡−1(−32)\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right) is
View answer and explanation
Correct option: B
−π3-\frac{\pi}{3}
Option B (−π3-\frac{\pi}{3}) is correct because the principal value range for sin⁡−1(x)\sin^{-1}(x) is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}], and sin⁡(−π3)=−32\sin(-\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}.
474 Easy
474 of 605
Understanding +4 | -1
If sin⁡−1x+sin⁡−1y=π2{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \frac{\pi }{2}, then cos⁡−1x+cos⁡−1y{\cos ^{ - 1}}x + {\cos ^{ - 1}}y is equal to
View answer and explanation
Correct option: A
π2\frac{\pi }{2}
Option A (π2\frac{\pi }{2}) is correct because if sin⁡−1x+sin⁡−1y=π2{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \frac{\pi }{2}, then cos⁡−1x+cos⁡−1y{\cos ^{ - 1}}x + {\cos ^{ - 1}}y can be derived using the identity sin⁡−1x+cos⁡−1x=π2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}.
475 Medium
475 of 605
Analyzing +4 | -1
The domain of the function Cos⁡−1(∣x∣−32)+1log⁡10(4−x)\operatorname{Cos}^{-1}\left(\frac{|x|-3}{2}\right)+\frac{1}{\log _{10}(4-x)} is
View answer and explanation
Correct option: A
[−5,−1]∪[1,3)∪(3,4)[-5,-1] \cup[1,3) \cup(3,4)
Option A ([-5,-1] ∪ [1,3) ∪ (3,4)) is correct because it represents the intersection of the domains of the inverse cosine and the logarithmic functions.
476 Medium
476 of 605
Application +4 | -1
The value of tan⁡{cos⁡−145+tan⁡−125}\tan \left\{\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{5}\right\} is λμ;(λ,μ)=1\frac{\lambda}{\mu} ;(\lambda, \mu)=1, then evaluate λ+19μ\frac{\lambda+19}{\mu}.
View answer and explanation
Correct option: A
3
Option A (3) is correct because the expression simplifies to 2314\frac{23}{14}, so λ=23\lambda = 23 and μ=14\mu = 14, and λ+19μ=3\frac{\lambda+19}{\mu} = 3.
477 Easy
477 of 605
Application +4 | -1
The sum of the two angles cot⁡−13{\cot ^{ - 1}}3 and cosec−15\mathrm{cose}{\mathrm{c}^{ - 1}}\sqrt 5, is
View answer and explanation
Correct option: C
π4\frac{\pi }{4}
Option C (π4\frac{\pi }{4}) is correct because by converting both inverse cotangent and inverse cosecant to inverse tangent, the sum simplifies to tan⁡−1(1)\tan^{-1}(1), which equals π4\frac{\pi}{4}.
478 Easy
478 of 605
Application +4 | -1
The Principal value of cos⁡−1(cos⁡8)−sin⁡−1sin⁡8\cos ^{-1}(\cos 8)-\sin ^{-1} \sin 8 is
View answer and explanation
Correct option: C
16−5π16-5 \pi
Option C (16−5π16-5 \pi) is correct because it involves correctly finding the principal values of cos⁡−1(cos⁡8)\cos^{-1}(\cos 8) and sin⁡−1(sin⁡8)\sin^{-1}(\sin 8) and then subtracting them.
479 Medium
479 of 605
Analyzing +4 | -1
Let f(x)=sin⁡−1(sin⁡x)f(x)=\sin ^{-1}(\sin x) and g(x)=cos⁡−1[x+1]+sin⁡−1[x]g(x)=\cos ^{-1}[x+1]+\sin ^{-1}[x]. Then the number of solution of f(x)+g(x)=4f(x)+g(x)=4, lies in the interval

(where [∗]\left[{ }^*\right] denotes the greatest integer function)
View answer and explanation
Correct option: C
(−2,1)(-2,1)
Option C, (−2,1)(-2,1), is correct because the domain of g(x) is [-1, 1) and within this domain, g(x) = 0, leading to no solution for f(x) + g(x) = 4.
480 Medium
480 of 605
Application +4 | -1
If tan⁡−1n,tan⁡−1(n+1)\tan ^{-1} \mathrm{n}, \tan ^{-1}(\mathrm{n}+1) and tan⁡−1(n+2),n∈N\tan ^{-1}(n+2), \mathrm{n} \in \mathbb{N}, are angles of a triangle, then find n\mathrm{n}.
View answer and explanation
Correct option: A
1
Option A (1) is correct because when n=1, the angles are tan-1(1), tan-1(2), and tan-1(3), which can be angles of a triangle.
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