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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Mixed difficulty

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61 Easy
61 of 605
Application +4 | -1
Sum of series S=tan⁡−1212+tan⁡−1222+tan⁡−1232+…S=\tan ^{-1} \frac{2}{1^2}+\tan ^{-1} \frac{2}{2^2}+\tan ^{-1} \frac{2}{3^2}+\ldots upto infinite terms, is :
View answer and explanation
Correct option: C
3π4\frac{3 \pi}{4}
The correct answer is C (3π4\frac{3 \pi}{4}) because the series can be written as a telescoping series using the arctangent subtraction formula, and the sum converges to 3π4\frac{3\pi}{4}.
62 Easy
62 of 605
Application +4 | -1
If ∑r = 150tan⁡−112r2=p\sum\limits_{r\, = \,1}^{50} {{{\tan }^{ - 1}}\frac{1}{{2{r^2}}} = p}, then the value of the tan⁡ p\tan \,p is
View answer and explanation
Correct option: B
5051\frac{{50}}{{51}}
Option B (5051\frac{{50}}{{51}}) is correct because it is the tangent of the sum of the arctangent series, which simplifies to a telescoping series.
63 Medium
63 of 605
Application +4 | -1
If asin⁡−1x−bcos⁡−1x=ca \sin ^{-1} x-b \cos ^{-1} x=c, then asin⁡−1x+a \sin ^{-1} x+ bcos⁡−1xb \cos ^{-1} x is equal to
View answer and explanation
Correct option: D
πab+c(a−b)a+b\frac{\pi a b+c(a-b)}{a+b}
Option D (πab+c(a−b)a+b\frac{\pi a b+c(a-b)}{a+b}) is correct because by solving the system of equations, we can find the value of the expression.
64 Medium
64 of 605
Application +4 | -1
If two angles of a Δ\Delta are tan⁡−112\tan ^{-1} \frac{1}{2} and tan⁡−11/3\tan ^{-1} 1 / 3, then find θ15\frac{\theta}{15} (where θ\theta is third angle of Δ\Delta in degree).
View answer and explanation
Correct option: A
9
Option A (9) is correct because the third angle is 135 degrees, and 135/15 = 9.
65 Hard
65 of 605
Application +4 | -1
If ∑i=110cos⁡−1xi=0\sum\limits_{{i=1}}^{10} \cos ^{-1} x_i=0, then ∑i=110xi\sum\limits_{{i=1}}^{10} x_i is
View answer and explanation
Correct option: B
10
Option B (10) is correct because each term in the summation must be zero, implying each xix_i is 1, and the sum of ten 1's is 10.
66 Easy
66 of 605
Application +4 | -1
sin⁡(12cos⁡−145)\sin \left( {\frac{1}{2}{{\cos }^{ - 1}}\frac{4}{5}} \right) is equal to
View answer and explanation
Correct option: B
110\frac{1}{{\sqrt {10} }}
Option B (110\frac{1}{{\sqrt {10} }}) is correct because using the half-angle formula for sine, the expression simplifies to 110\frac{1}{{\sqrt {10} }}.
67 Medium
67 of 605
Application +4 | -1
The value of sin⁡−1(sin⁡10)\sin ^{-1}(\sin 10) is -
View answer and explanation
Correct option: C
3π−103 \pi-10
Option C (3π−103 \pi - 10) is correct because the range of sin⁡−1(x)\sin^{-1}(x) is [−π/2,π/2][-\pi/2, \pi/2], and we need to find an angle in this range that has the same sine value as 10.
68 Easy
68 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 it is the correct simplification of the given expression within the specified range of x, using trigonometric identities.
69 Easy
69 of 605
Application +4 | -1
If 3tan⁡−112+3−tan⁡−11x=tan⁡−1133 \tan ^{-1} \frac{1}{2+\sqrt{3}}-\tan ^{-1} \frac{1}{x}=\tan ^{-1} \frac{1}{3} then x=x=
View answer and explanation
Correct option: A
2
Option A (2) is correct because by simplifying the inverse trigonometric equation using tan⁡15∘=2−3\tan 15^{\circ}=2-\sqrt{3} and the formula for tan⁡−1x−tan⁡−1y\tan^{-1}x - \tan^{-1}y, we find that x=2x=2.
70 Medium
70 of 605
Analyzing +4 | -1
If (x−1)(x2+2)>0(x-1)\left(x^2+2\right) > 0, then the value of sin⁡(12tan⁡−1(2x1−x2)−tan⁡−1x)\sin \left(\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^2}\right)-\tan ^{-1} x\right) is
View answer and explanation
Correct option: B
−1-1
Option B (-1) is correct because, given the condition x > 1, the expression simplifies to sin(-π/2), which equals -1.
71 Easy
71 of 605
Application +4 | -1
sin⁡−1(35)+tan⁡−1(17)={\sin ^{ - 1}}\left( {\frac{3}{5}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right) =
View answer and explanation
Correct option: A
π4\frac{\pi }{4}
Option A (π4\frac{\pi }{4}) is correct because the sum of the inverse trigonometric functions simplifies to tan⁡−1(1)\tan^{-1}(1), which equals π4\frac{\pi }{4}.
72 Medium
72 of 605
Application +4 | -1
If cos⁡−1p+cos⁡−11−p+cos⁡−11−q=3π4\cos ^{-1} \sqrt{p}+\cos ^{-1} \sqrt{1-p}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}, then the value of qq is
View answer and explanation
Correct option: A
0.5
Option A (0.5) is correct because by using trigonometric identities and solving the equation, the value of q is found to be 1/2.
73 Easy
73 of 605
Application +4 | -1
If x∈(  −∞,  1)x \in \left( {\; - \infty ,\;1} \right), then tan⁡−1(2x1−x2){\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right) equals
View answer and explanation
Correct option: C
π+2tan⁡−1x\pi + 2{\tan ^{ - 1}}x
Option C (π+2tan⁡−1x\pi + 2{\tan ^{ - 1}}x) is correct because when x∈(−∞,−1)x \in ( - \infty , - 1), the expression tan⁡−1(2x1−x2){\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right) simplifies to π+2tan⁡−1x\pi + 2{\tan ^{ - 1}}x due to the range of the arctangent function.
74 Easy
74 of 605
Analyzing +4 | -1
The number of real solutions of tan⁡−1x(x+1)+sin⁡−1x2+x+1=π2{\tan ^{ - 1}}\sqrt {x\left( {x + 1} \right)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = \frac{\pi }{2} is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the equation simplifies to x2+x=0{x^2} + x = 0, which has two real solutions: x=0x = 0 and x=−1x = -1.
75 Easy
75 of 605
Application +4 | -1
2sin⁡−1(−1/2)+3cos⁡−1(−3/2)+4sin⁡−1(1/2)2 \sin ^{-1}(-1 / 2)+3 \cos ^{-1}(-\sqrt{3} / 2)+4 \sin ^{-1}(1 / \sqrt{2}) +5tan⁡−1(−1)+5 \tan ^{-1}(-1) is equal to (considering the principal values)
View answer and explanation
Correct option: C
23π/1223 \pi / 12
Option C (23π/1223\pi/12) is correct because it is the result of correctly evaluating each inverse trigonometric term and summing them.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
76 Easy
76 of 605
Application +4 | -1
The value of sin⁡(π3−sin⁡−1(−12))\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right) is
View answer and explanation
Correct option: D
1
Option D (1) is correct because sin⁡(π3−sin⁡−1(−12))\sin(\frac{\pi}{3} - \sin^{-1}(-\frac{1}{2})) simplifies to sin⁡(π3+π6)\sin(\frac{\pi}{3} + \frac{\pi}{6}), which equals sin⁡(π2)\sin(\frac{\pi}{2}) or 1.
77 Easy
77 of 605
Application +4 | -1
The value of sin⁡−1(cos⁡33π5){\sin ^{ - 1}}\left( {\cos \frac{{33\pi }}{5}} \right) is
View answer and explanation
Correct option: D
−π10- \frac{\pi }{{10}}
Option D (−π10- \frac{\pi }{{10}}) is correct because by simplifying the expression inside the inverse sine function using trigonometric identities, we find that the value is −π10- \frac{\pi }{{10}}.
78 Easy
78 of 605
Application +4 | -1
The value of sin (cot⁡−1x)({\cot ^{ - 1}}x) is
View answer and explanation
Correct option: D
(1+x2)−1/2{\left( {1 + {x^2}} \right)^{ - 1/2}}
Option D ((1+x2)−1/2{\left( {1 + {x^2}} \right)^{ - 1/2}}) is correct because by letting cot⁡−1x=θ{\cot ^{ - 1}}x = \mathrm{\theta }, we can find sin⁡θ\sin{\theta} in terms of x.
79 Easy
79 of 605
Application +4 | -1
sin⁡[2sin⁡−1(3/5)+2sin⁡−1(4/5)]\sin \left[2 \sin ^{-1}(3 / 5)+2 \sin ^{-1}(4 / 5)\right] is equal to
View answer and explanation
Correct option: B
336/625336 / 625
Option B (336/625336 / 625) is correct because by using the double angle formula for sine and simplifying, we arrive at the answer 336/625336/625.
80 Easy
80 of 605
Application +4 | -1
If y=tan⁡−11−x1+x,0≤x≤1y=\tan ^{-1} \frac{1-x}{1+x}, 0 \leq x \leq 1, then
View answer and explanation
Correct option: B
0≤y≤π/40 \leq y \leq \pi / 4
The correct answer is B (0≤y≤π/40 \leq y \leq \pi / 4) because substituting x=tan⁡θx = \tan \theta and simplifying the expression leads to this range for y.
81 Easy
81 of 605
Application +4 | -1
If (tan⁡−1x)2+(cot⁡−1x)2=5π28{\left( {{{\tan }^{ - 1}}x} \right)^2} + {\left( {{{\cot }^{ - 1}}x} \right)^2} = \frac{{5{\pi ^2}}}{8}, then xx equals
View answer and explanation
Correct option: A
−1- 1
Option A (−1-1) is correct because by using the identity tan⁡−1x+cot⁡−1x=π2{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2} and solving the resulting quadratic equation, we find that x=−1x = -1.
82 Easy
82 of 605
Application +4 | -1
cos⁡−1(12(cos⁡7π5−sin⁡2π5))\cos ^{-1}\left(\frac{1}{\sqrt{2}}\left(\cos \frac{7 \pi}{5}-\sin \frac{2 \pi}{5}\right)\right) is equal to
View answer and explanation
Correct option: D
17π20\frac{17 \pi}{20}
Option D (17π20\frac{{17 \pi}}{{20}}) is correct because the expression simplifies to cos⁡(17π20)\cos(\frac{{17 \pi}}{{20}}), and the inverse cosine function returns 17π20\frac{{17 \pi}}{{20}}.
83 Easy
83 of 605
Analyzing +4 | -1
If x1,x2,  x3,  x4{x_1},{x_2},\;{x_3},\;{x_4} are the roots of the equation x4−x3sin⁡2β−xcos⁡β−sin⁡β=0,  {x^4} - {x^3}\sin 2\mathrm{\beta } - x\cos \mathrm{\beta } - \sin \mathrm{\beta } = 0,\; then tan⁡−1+tan⁡−1x2+tan⁡−1x3+tan⁡−1x4{\tan ^{ - 1}} + {\tan ^{ - 1}}{x_2} + {\tan ^{ - 1}}{x_3} + {\tan ^{ - 1}}{x_4} is equal to
View answer and explanation
Correct option: B
π2−β\frac{\pi }{2} - \mathrm{\beta }
Option B (π2−β\frac{\pi }{2} - \mathrm{\beta }) is correct because by using Vieta's formulas and the arctangent addition formula, we can simplify the expression to π2−β\frac{\pi }{2} - \mathrm{\beta }.
84 Medium
84 of 605
Application +4 | -1
cosec⁡−1(cos⁡x)\operatorname{cosec}^{-1}(\cos x) is real if
View answer and explanation
Correct option: D
xx is integer multiple of π\pi
Option D (xx is integer multiple of π\pi) is correct because for cosec⁡−1(cos⁡x)\operatorname{cosec}^{-1}(\cos x) to be real, ∣cos⁡x∣|\cos x| must be equal to 1, which occurs when xx is an integer multiple of π\pi.
85 Medium
85 of 605
Application +4 | -1
The value of tan⁡(tan⁡−11+tan⁡−12+tan⁡−13)=\tan \left(\tan ^{-1} 1+\tan ^{-1} 2+\tan ^{-1} 3\right)=
View answer and explanation
Correct option: A
0
Option A (0) is correct because the sum of the inverse tangents simplifies to 0.
86 Easy
86 of 605
Application +4 | -1
tan⁡−1xy−tan⁡−1x−yx+y  {\tan ^{ - 1}}\frac{x}{y} - {\tan ^{ - 1}}\frac{{x - y}}{{x + y}}\; is equal to

(where x<y>0)x < y > 0)
View answer and explanation
Correct option: B
π4\frac{\pi }{4}
Option B (π4\frac{\pi }{4}) is correct because by simplifying the expression using trigonometric identities, we find that it equals π4\frac{\pi}{4}.
87 Easy
87 of 605
Application +4 | -1
The value of tan⁡−1(1)+tan⁡−1(0)+tan⁡−1(2)+tan⁡−1(3){\tan ^{ - 1}}\left( 1 \right) + {\tan ^{ - 1}}\left( 0 \right) + {\tan ^{ - 1}}\left( 2 \right) + {\tan ^{ - 1}}\left( 3 \right) is equal to
View answer and explanation
Correct option: A
π\pi
Option A (π\pi) is correct because simplifying the expression tan⁡−1(1)+tan⁡−1(0)+tan⁡−1(2)+tan⁡−1(3){\tan ^{ - 1}}(1) + {\tan ^{ - 1}}(0) + {\tan ^{ - 1}}(2) + {\tan ^{ - 1}}(3) using the inverse tangent addition formula results in π\pi.
88 Medium
88 of 605
Application +4 | -1
If sin⁡−1x+sin⁡−1y+sin⁡−1z=π\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\pi, then x1−x2+y1−y2+z1−z2x \sqrt{1-x^2}+y \sqrt{1-y^2}+z \sqrt{1-z^2} equals
View answer and explanation
Correct option: D
2xyz2 x y z
Option D (2xyz2xyz) is correct because, given the condition, the expression simplifies to 2xyz2xyz using trigonometric identities.
89 Medium
89 of 605
Application +4 | -1
If 12sin⁡−1(3sin⁡2θ5+4cos⁡2θ)=π4\frac{1}{2} \sin ^{-1}\left(\frac{3 \sin 2 \theta}{5+4 \cos 2 \theta}\right)=\frac{\pi}{4}, then tan⁡θ\tan \theta is equal to
View answer and explanation
Correct option: B
3
Option B (3) is correct because solving the equation after applying the sine function and simplifying leads to tan⁡θ=3\tan \theta = 3.
90 Medium
90 of 605
Analyzing +4 | -1
sin⁡−112+sin⁡−12−16+sin⁡−13−212+sin⁡−12−320+sin⁡−15−230+…∞\sin ^{-1} \frac{1}{\sqrt{2}}+\sin ^{-1} \frac{\sqrt{2}-1}{\sqrt{6}}+\sin ^{-1} \frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+\sin ^{-1} \frac{2-\sqrt{3}}{\sqrt{20}}+\sin ^{-1} \frac{\sqrt{5}-2}{\sqrt{30}}+\ldots \infty
View answer and explanation
Correct option: B
π2\frac{\pi}{2}
The correct answer is π2\frac{\pi}{2} because the given series telescopes to π2\frac{\pi}{2} as nn approaches infinity.
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