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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Mixed difficulty

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Showing 301-330 of 605 questions

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301 Easy
301 of 605
Analyzing +4 | -1
The equation 2cos⁡−1x+sin⁡−1x=11π62{\cos ^{ - 1}}x + {\sin ^{ - 1}}x = \frac{{11\pi }}{6} has
View answer and explanation
Correct option: A
No solution
Option A (tan⁡−1(n2+nn2+n+2){\tan ^{ - 1}}\left( {\frac{{{n^2} + n}}{{{n^2} + n + 2}}} \right)) is correct because the summation telescopes, resulting in the given expression.
302 Medium
302 of 605
Application +4 | -1
If sin⁡−1(x5)+cosec⁡−1(54)=π2\sin ^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}, then the values of x\mathrm{x} is
View answer and explanation
Correct option: D
3
Option D (3) is correct because using the identity sin⁡−1x+csc⁡−1y=π2\sin^{-1}x + \csc^{-1}y = \frac{{\pi}}{{2}} and solving for x yields the value 3.
303 Easy
303 of 605
Application +4 | -1
If 12≤x≤1\frac{1}{2} \le x \le 1, then sin⁡−1(3x−4x3){\sin ^{ - 1}}\left( {3x - 4{x^3}} \right) equals
View answer and explanation
Correct option: B
π−3sin⁡−1x\pi - 3{\sin ^{ - 1}}x
Option B (π−3sin⁡−1x\pi - 3{\sin ^{ - 1}}x) is correct because within the given range of x, the expression sin⁡−1(3x−4x3){\sin ^{ - 1}}(3x - 4{x^3}) simplifies to π−3sin⁡−1x\pi - 3{\sin ^{ - 1}}x.
304 Easy
304 of 605
Application +4 | -1
If sin⁡−1x+sin⁡−1y+sin⁡−1z=3π2{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \frac{{3\pi }}{2}, then the value of x100+y100+z100−9x101+y101+z101{x^{100}} + {y^{100}} + {z^{100}} - \frac{9}{{{x^{101}} + {y^{101}} + {z^{101}}}}, is
View answer and explanation
Correct option: A
0
Option A is correct because when sin⁡−1x+sin⁡−1y+sin⁡−1z=3π2{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \frac{{3\pi }}{2}, it implies x=y=z=1x=y=z=1, and therefore x100+y100+z100−9x101+y101+z101=0{x^{100}} + {y^{100}} + {z^{100}} - \frac{9}{{{x^{101}} + {y^{101}} + {z^{101}}}} = 0.
305 Easy
305 of 605
Analyzing +4 | -1
Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation cos⁡−1(x)−2sin⁡−1(x)=cos⁡−1(2x)\cos ^{-1}(x)-2 \sin ^{-1}(x)=\cos ^{-1}(2 x) is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because the sum of the solutions to the given equation is 0, obtained by solving the equation and summing the valid solutions.
306 Medium
306 of 605
Application +4 | -1
If α∈(−3π2,−π)\alpha \in\left(-\frac{3 \pi}{2},-\pi\right), then the value of tan⁡−1(cot⁡α)−\tan ^{-1}(\cot \alpha)- cot⁡−1(tan⁡α)+sin⁡−1(sin⁡α)+cos⁡−1(cos⁡α)\cot ^{-1}(\tan \alpha)+\sin ^{-1}(\sin \alpha)+\cos ^{-1}(\cos \alpha) is equal to:
View answer and explanation
Correct option: C
0
Option C (0) is correct because the terms cancel out based on the given range of alpha and properties of inverse trigonometric functions.
307 Easy
307 of 605
Application +4 | -1
The value of sin⁡[2cos⁡−153]\sin \left[ {2{{\cos }^{ - 1}}\frac{{\sqrt 5 }}{3}} \right] is
View answer and explanation
Correct option: C
459\frac{{4\sqrt 5 }}{9}
Option C (459\frac{{4\sqrt 5 }}{{9}}) is correct because it is the result of simplifying the given trigonometric expression using inverse trigonometric identities.
308 Medium
308 of 605
Application +4 | -1
The largest integral value of x{x} satisfying the inequality (tan⁡−1(x))2−4(tan⁡−1(x))+3>0\left(\tan ^{-1}(x)\right)^2-4\left(\tan ^{-1}(x)\right)+3>0 is:
View answer and explanation
Correct option: B
1
Option B (1) is correct because solving the inequality for arctangent and then finding the largest integer value of x that satisfies the condition.
309 Easy
309 of 605
Application +4 | -1
If cos⁡−1x+cos⁡−1y+cos⁡−1z=π,{\cos ^{ - 1}}x + {\cos ^{ - 1}}y + {\cos ^{ - 1}}z = \pi , then
View answer and explanation
Correct option: C
x2+y2+z2=1−2xyz{x^2} + {y^2} + {z^2} = 1 - 2xyz
Option D is correct because the given condition leads to the identity x2+y2+z2+2xyz=1{x^2} + {y^2} + {z^2} + 2xyz = 1, which can be rearranged to x2+y2+z2=1−2xyz{x^2} + {y^2} + {z^2} = 1 - 2xyz.
310 Easy
310 of 605
Application +4 | -1
The number of triplets (x,  y,  z)\left( {x,\;y,\;z} \right) satisfying sin⁡−1x+cos⁡−1y+sin⁡−1z=2π,  {\sin ^{ - 1}}x + {\cos ^{ - 1}}y + {\sin ^{ - 1}}z = 2\pi ,\; is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the equation is satisfied only when x = 1, y = -1, and z = 1, resulting in a single triplet.
311 Easy
311 of 605
Application +4 | -1
If −1≤x≤0- 1 \le x \le 0, then cos⁡−1(2x2−1){\cos ^{ - 1}}\left( {2{x^2} - 1} \right) equals
View answer and explanation
Correct option: C
2π−2cos⁡−1x2\pi - 2{\cos ^{ - 1}}x
Option C (2π−2cos⁡−1x2\pi - 2{\cos ^{ - 1}}x) is correct because under the given condition −1≤x≤0-1 \le x \le 0, cos⁡−1(2x2−1){\cos ^{ - 1}}(2x^2 - 1) simplifies to 2π−2cos⁡−1x2\pi - 2{\cos ^{ - 1}}x.
312 Easy
312 of 605
Application +4 | -1
If tan⁡−1x−1x+2+tan⁡−1x+1x+2=π4,{\tan ^{ - 1}}\frac{{x - 1}}{{x + 2}} + {\tan ^{ - 1}}\frac{{x + 1}}{{x + 2}} = \frac{\pi }{4}, then xx is equal to
View answer and explanation
Correct option: C
±52\pm \sqrt {\frac{5}{2}}
Option C (±52\pm \sqrt {\frac{5}{2}}) is correct because solving the given inverse tangent equation using the tangent addition formula results in x=±52x = \pm \sqrt {\frac{5}{2}}.
313 Medium
313 of 605
Application +4 | -1
tan⁡{2tan⁡−1(15)−π4}\tan \left\{2 \tan ^{-1}\left(\frac{1}{5}\right)-\frac{\pi}{4}\right\} is equal to
View answer and explanation
Correct option: C
−717-\frac{7}{17}
Option C (-7/17) is correct because it is the result of applying the tangent addition and double angle formulas to the given expression.
314 Easy
314 of 605
Application +4 | -1
sin⁡(2sin⁡−10.8)\sin \left(2 \sin ^{-1} 0.8\right) is equal to
View answer and explanation
Correct option: A
0.96
Option A (0.96) is correct because it is the result of applying the double angle formula for sine, sin⁡(2θ)=2sin⁡(θ)cos⁡(θ)\sin(2\theta) = 2\sin(\theta)\cos(\theta), with sin⁡(θ)=0.8\sin(\theta) = 0.8.
315 Medium
315 of 605
Application +4 | -1
If ∑i=12nsin⁡−1xi=nπ\sum\limits_{{i=1}}^{2 n} \sin ^{-1} x_i=n \pi, then ∑i=12nxi\sum\limits_{{i=1}}^{2 n} x_i is
View answer and explanation
Correct option: B
2n2 n
Option B (2n) is correct because the sum of inverse sines equaling nπ implies each term must be π/2, leading to each x_i being 1.

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316 Easy
316 of 605
Application +4 | -1
If cos⁡−1x−cos⁡−1y2=α\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha, then 4x2−4xycos⁡α+y24 x^{2}-4 x y \cos \alpha+y^{2} is equal to
View answer and explanation
Correct option: C
4sin⁡2α4 \sin ^{2} \alpha
Option C (4sin⁡2α4 \sin ^{2} \alpha) is correct because after applying the cosine subtraction formula and simplifying, the expression equals 4sin⁡2α4 \sin^2 \alpha.
317 Easy
317 of 605
Understanding +4 | -1
sin⁡−1x+sin⁡−11x+cos⁡−1x+cos⁡−11x{\sin ^{ - 1}}x + {\sin ^{ - 1}}\frac{1}{x} + {\cos ^{ - 1}}x + {\cos ^{ - 1}}\frac{1}{x} is equal to
View answer and explanation
Correct option: A
π\pi
Option A (π\pi) is correct because using the identity sin⁡−1x+cos⁡−1x=π2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}, the expression simplifies to π2+π2=π\frac{\pi }{2} + \frac{\pi }{2} = \pi.
318 Easy
318 of 605
Application +4 | -1
If −13<x<13- \frac{1}{{\sqrt 3 }} < x < \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: A
3tan⁡−1x3{\tan ^{ - 1}}x
Option A (3tan⁡−1x3{\tan ^{ - 1}}x) is correct because, within the given range, the expression simplifies directly using the triple angle formula for tangent.
319 Easy
319 of 605
Application +4 | -1
The value of tan⁡(2tan⁡−1(35)+sin⁡−1(513))\tan \left(2 \tan ^{-1}\left(\frac{3}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)\right) is equal to:
View answer and explanation
Correct option: A
22021\frac{220}{21}
Option A (22021\frac{{220}}{{21}}) is correct because it is the result of applying the tangent addition formula after converting the inverse sine to an inverse tangent.
320 Medium
320 of 605
Analyzing +4 | -1
∑m=1ntan⁡−1(2mm4+m2+2)  \sum \limits_{m = 1}^n {\tan ^{ - 1}}\left( {\frac{{2m}}{{{m^4} + {m^2} + 2}}} \right)\; is equal to
View answer and explanation
Correct option: A
tan⁡−1(n2+nn2+n+2){\tan ^{ - 1}}\left( {\frac{{{n^2} + n}}{{{n^2} + n + 2}}} \right)
Option A (tan⁡−1(n2+nn2+n+2){\tan ^{ - 1}}\left( {\frac{{{{{n^2} + n}}}}{{{{{n^2} + n + 2}}}}} \right)) is correct because the sum telescopes, leaving a simplified expression involving nn.
321 Easy
321 of 605
Analyzing +4 | -1
If y=tan⁡−1(sec⁡x3−tan⁡x3),π2<x3<3π2,y = {\tan ^{ - 1}}(\sec {x^3} - \tan {x^3}),\frac{\pi }{2} < {x^3} < \frac{{3\pi }}{2}, then
View answer and explanation
Correct option: B
x2y′′−6y+3π2=0{x^2}y'' - 6y + \frac{{3\pi }}{2} = 0
Option B (x2y′′−6y+3π2=0{x^2}y'' - 6y + \frac{{3\pi }}{{2}} = 0) is correct because after simplifying the given equation and differentiating twice, it satisfies the given relation.
322 Easy
322 of 605
Application +4 | -1
cos⁡−1(−12)−2sin⁡−1(12)+3cos⁡−1(−12)    −4tan⁡−1(−1){\cos ^{ - 1}}\left( {\frac{{ - 1}}{2}} \right) - 2{\sin ^{ - 1}}\left( {\frac{1}{2}} \right) + 3{\cos ^{ - 1}}\left( {\frac{{ - 1}}{{\sqrt 2 }}} \right)\;\; - 4{\tan ^{ - 1}}\left( { - 1} \right) equals
View answer and explanation
Correct option: C
47π12\frac{{47\pi }}{{12}}
Option D (43π12\frac{{43\pi }}{{12}}) is correct because by evaluating each inverse trigonometric function and simplifying the expression, the result is 43π12\frac{{43\pi}}{{12}}.
323 Medium
323 of 605
Application +4 | -1
The value of cos⁡[12cos⁡−1{cos⁡(sin⁡−1638)}]\cos \left[ {\frac{1}{2}{{\cos }^{ - 1}}\left\{ {\cos \left( {{{\sin }^{ - 1}}\frac{{\sqrt {63} }}{8}} \right)} \right\}} \right], is
View answer and explanation
Correct option: C
34\frac{3}{4}
Option C (3/4) is correct because by simplifying the given expression using trigonometric identities, we arrive at the value 3/4.
324 Medium
324 of 605
Application +4 | -1
A value of tan⁡−1(sin⁡(cos⁡−1(23)))\tan ^{-1}\left(\sin \left(\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)\right)\right) is
View answer and explanation
Correct option: D
π6\frac{\pi}{6}
The correct answer is D (π6\frac{\pi}{6}) because it involves simplifying the expression using trigonometric identities and inverse trigonometric functions.
325 Medium
325 of 605
Application +4 | -1
The value of sec⁡−1[cosec⁡{cot⁡(cot⁡−1(3π4))}]\sec ^{-1}\left[\operatorname{cosec}\left\{\cot \left(\cot ^{-1}\left(\frac{3 \pi}{4}\right)\right)\right\}\right] is
View answer and explanation
Correct option: C
π4\frac{\pi}{4}
Option C (π/4) is correct because simplifying the expression inside out leads to sec⁻¹(cosec(cot(cot⁻¹(3π/4)))) = sec⁻¹(cosec(3π/4)) = sec⁻¹(√2) = π/4.
326 Medium
326 of 605
Application +4 | -1
Number of values of xx satisfying sin⁡−1(1−x)−2sin⁡−1x=π/2\sin ^{-1}(1-x)-2 \sin ^{-1} x=\pi / 2 is
View answer and explanation
Correct option: B
1
The correct answer is B (1) because after solving the equation, only one value of x satisfies the original equation.
327 Easy
327 of 605
Application +4 | -1
If x<0x < 0, then tan⁡−1(1x){\tan ^{ - 1}}\left( {\frac{1}{x}} \right) equals
View answer and explanation
Correct option: C
−π+cot⁡−1x- \pi + {\cot ^{ - 1}}x
Option C (−π+cot⁡−1x- \pi + {\cot ^{ - 1}}x) is correct because when x<0x < 0, the range of tan⁡−1(1x){\tan ^{ - 1}}(\frac{1}{x}) is (−π2,0)(-\frac{\pi}{2}, 0), and the correct transformation using cot⁡−1x{\cot ^{ - 1}}x accounts for this.
328 Easy
328 of 605
Application +4 | -1
If 6sin⁡−1(x2−6x+8.5)=π6 \sin ^{-1}\left(x^2-6 x+8.5\right)=\pi, then xx is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because it is one of the solutions obtained after solving the trigonometric equation and lies within the domain of the inverse sine function.
329 Easy
329 of 605
Application +4 | -1
If tan⁡−1x−1x+2+tan⁡−1x+1x+2=π4,{\tan ^{ - 1}}\frac{{x - 1}}{{x + 2}} + {\tan ^{ - 1}}\frac{{x + 1}}{{x + 2}} = \frac{\pi }{4}, then xx is equal to
View answer and explanation
Correct option: C
±52\pm \sqrt {\frac{5}{2}}
Option C (±52\pm \sqrt {\frac{5}{2}}) is correct because solving the given equation using the tangent addition formula leads to this solution for x.
330 Easy
330 of 605
Application +4 | -1
If x≥1x \geq 1, then 2tan⁡−1x+sin⁡−12x1+x22 \tan ^{-1} x+\sin ^{-1} \frac{2 x}{1+x^2} is equal to
View answer and explanation
Correct option: B
π\pi
Option B (π\pi) is correct because when x ≥ 1, the expression 2tan⁻¹x + sin⁻¹(2x/(1+x²)) simplifies to π.
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