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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Easy

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 31-60 of 365 questions
31 Easy
31 of 365
Application +4 | -1
If tan⁡−1x+tan⁡−1y=π4{\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \frac{\pi }{4}, then
View answer and explanation
Correct option: A
x+y+xy=1x + y + xy = 1
Option A (x+y+xy=1x + y + xy = 1) is correct because it is derived by applying the tangent addition formula to the given equation and simplifying.
32 Easy
32 of 365
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}.
33 Easy
33 of 365
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.
34 Easy
34 of 365
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} }}.
35 Easy
35 of 365
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.
36 Easy
36 of 365
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.
37 Easy
37 of 365
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}.
38 Easy
38 of 365
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.
39 Easy
39 of 365
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.
40 Easy
40 of 365
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.
41 Easy
41 of 365
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.
42 Easy
42 of 365
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}}.
43 Easy
43 of 365
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.
44 Easy
44 of 365
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.
45 Easy
45 of 365
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.

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46 Easy
46 of 365
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.
47 Easy
47 of 365
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}}.
48 Easy
48 of 365
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 }.
49 Easy
49 of 365
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}.
50 Easy
50 of 365
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.
51 Easy
51 of 365
Application +4 | -1
Two angles of a triangle are cot⁡−12{\cot ^{ - 1}}2 and cot⁡−13{\cot ^{ - 1}}3. Then , the third angle is
View answer and explanation
Correct option: B
3  π4\frac{{3\;\pi }}{4}
Option B (3  π4\frac{{3\;\pi }}{4}) is correct because the sum of the given angles is π4\frac{\pi}{4}, and the third angle is π−π4=3π4\pi - \frac{\pi}{4} = \frac{3\pi}{4}.
52 Easy
52 of 365
Application +4 | -1
If cot⁡(cos⁡−1x)=sec⁡(tan⁡−1ab2−a2)\cot \left( {{{\cos }^{ - 1}}x} \right) = \sec \left( {{{\tan }^{ - 1}}\frac{a}{{\sqrt {{b^2} - {a^2}} }}} \right), then xx is equal to
View answer and explanation
Correct option: A
b2b2−a2\frac{b}{{\sqrt {2{b^2} - {a^2}} }}
Option A (b2b2−a2\frac{b}{{\sqrt {2{b^2} - {a^2}} }}) is correct because by converting both sides to cotangent and secant expressions, we can solve for x.
53 Easy
53 of 365
Application +4 | -1
If sin⁡−1(3x)+sin⁡−1(4x)=π2{\sin ^{ - 1}}\left( {\frac{3}{x}} \right) + {\sin ^{ - 1}}\left( {\frac{4}{x}} \right) = \frac{\pi }{2}, then xx is equal to
View answer and explanation
Correct option: B
5
Option B (5) is correct because solving the equation using trigonometric identities and squaring both sides leads to x = 5.
54 Easy
54 of 365
Application +4 | -1
cos⁡−1(12)+2sin⁡−1(12){\cos ^{ - 1}}\left( {\frac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\frac{1}{2}} \right) is equal to
View answer and explanation
Correct option: C
2π3\frac{{2\pi }}{3}
Option C (2π3\frac{{2\pi }}{{3}}) is correct because it is the result of directly substituting the known values of the inverse trigonometric functions and simplifying.
55 Easy
55 of 365
Application +4 | -1
If 2 tan⁡−1(cos  x)=tan⁡−1(2  cosec  x),  {\tan ^{ - 1}}\left( {\mathrm{cos}\;x} \right) = {\tan ^{ - 1}}\left( {2\;\mathrm{cosec}\;x} \right),\; then the value of xx is
View answer and explanation
Correct option: B
π4\frac{\pi }{4}
Option B (π4\frac{\pi}{4}) is correct because solving the given trigonometric equation involving inverse tangents leads to x=π4x = \frac{\pi}{4}.
56 Easy
56 of 365
Application +4 | -1
If tan⁡(cos⁡−1x)=sin⁡cot⁡−1(1/2)\tan \left(\cos ^{-1} x\right)=\sin \cot ^{-1}(1 / 2), then the value of xx is
View answer and explanation
Correct option: C
5/3\sqrt{5} / 3
Option C (5/3\sqrt{5}/3) is correct because it is the solution obtained by converting the inverse trigonometric functions to tangent and sine, respectively, and solving for x.
57 Easy
57 of 365
Application +4 | -1
The principal value of tan⁡−1(cot⁡43π/4)\tan ^{-1}(\cot 43 \pi / 4) is
View answer and explanation
Correct option: D
−π/4-\pi / 4
Option D (−π/4- \pi / 4) is correct because by simplifying the expression using trigonometric identities, we find the principal value of the inverse tangent.
58 Easy
58 of 365
Application +4 | -1
If sin⁡−1(x/5)+cosec⁡−1(5/4)=π/2\sin ^{-1}(x / 5)+\operatorname{cosec}^{-1}(5 / 4)=\pi / 2 then a values of xx is
View answer and explanation
Correct option: B
3
Option B (3) is correct because using the identity cosec⁡−1(x)=sin⁡−1(1/x)\operatorname{cosec}^{-1}(x) = \sin^{-1}(1/x) and sin⁡−1x+cos⁡−1x=π/2\sin^{-1}x + \cos^{-1}x = \pi/2, we can solve for xx.
59 Easy
59 of 365
Application +4 | -1
tan⁡−1(14)+tan⁡−1(29)={\tan ^{ - 1}}\left( {\frac{1}{4}} \right) + {\tan ^{ - 1}}\left( {\frac{2}{9}} \right) =
View answer and explanation
Correct option: D
tan⁡−1(12){\tan ^{ - 1}}\left( {\frac{1}{2}} \right)
Option D (tan⁡−1(12){\tan ^{ - 1}}\left( {\frac{1}{2}} \right)) is correct because it is the simplified value of the given expression after applying the tangent addition formula.
60 Easy
60 of 365
Application +4 | -1
If x∈[−1,  1]x \in \left[ { - 1,\;1} \right], then sin⁡−1(2x1+x2){\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right) equals
View answer and explanation
Correct option: A
2tan⁡−1x2{\tan ^{ - 1}}x
Option A (2tan⁡−1x2{\tan ^{ - 1}}x) is correct because by substituting x=tan⁡θx = \tan \theta, the expression simplifies to sin⁡−1(sin⁡2θ){\sin ^{ - 1}}(\sin 2\theta) which equals 2θ2\theta under the given condition.
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