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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Easy

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121 Easy
121 of 365
Analyzing +4 | -1
If x,y and z are in A.P and tan⁡−1x,  tan⁡−1y    and   tan⁡−1z{\tan ^{ - 1}}x,\,\,{\tan ^{ - 1}}y\,\,\,\,\mathrm{and}\,\,\,{\tan ^{ - 1}}z are also in AP, then
View answer and explanation
Correct option: A
x=y=zx = y = z
Option A (x=y=zx = y = z) is correct because if x, y, z are in both A.P. and their arctangents are in A.P., it implies they must be equal.
122 Easy
122 of 365
Application +4 | -1
A value of xx for which sin⁡(cot⁡−1(1+x))=cos⁡(tan⁡−1x)\sin \left(\cot ^{-1}(1+x)\right)=\cos \left(\tan ^{-1} x\right) is
View answer and explanation
Correct option: A
−(1/2)-(1 / 2)
Option A (-(1 / 2)) is correct because substituting x = -(1/2) into the equation and using trigonometric identities leads to a valid solution.
123 Easy
123 of 365
Application +4 | -1
If θ=tan⁡−1a,ϕ\mathrm{\theta } = {\tan ^{ - 1}}a,\phi =tan⁡−1b  and  ab{\tan ^{ - 1}}b\;\mathrm{and}\;ab=-1,then (θ−ϕ)\mathrm{\theta } - \phi ) is equal to
View answer and explanation
Correct option: C
π2\frac{\pi }{2}
Option C (π2\frac{\pi}{2}) is correct because given the condition ab = -1, the difference between the inverse tangents results in π2\frac{\pi}{2}.
124 Easy
124 of 365
Application +4 | -1
4tan⁡−115−tan⁡−112394{\tan ^{ - 1}}\frac{1}{5} - {\tan ^{ - 1}}\frac{1}{{239}} is equal to
View answer and explanation
Correct option: D
π/4\pi /4
Option D (π/4\pi /4) is correct because using the tangent addition formula repeatedly simplifies the expression to tan⁡−1(1){\tan ^{ - 1}}(1), which equals π/4\pi /4.
125 Easy
125 of 365
Application +4 | -1
tan⁡2π5−tan⁡π15−3tan⁡2  π5tan⁡π15\tan \frac{{2\pi }}{5} - \tan \frac{\pi }{{15}} - \sqrt 3 \tan \frac{{2\;\pi }}{5}\tan \frac{\pi }{{15}} is equal to
View answer and explanation
Correct option: D
3\sqrt 3
Option A (235\frac{2}{{3\sqrt 5 }}) is correct because the expression simplifies to tan⁡(sin⁡−1(27))\tan({\sin ^{ - 1}}(\frac{2}{7})), which equals 235\frac{2}{{3\sqrt 5 }}.
126 Easy
126 of 365
Application +4 | -1
The value of xx, where x>0x > 0 and tan⁡{sec⁡−1(1x)}=sin⁡(tan⁡−12)\tan \left\{ {{{\sec }^{ - 1}}\left( {\frac{1}{x}} \right)} \right\} = \sin \left( {{{\tan }^{ - 1}}2} \right) is
View answer and explanation
Correct option: B
53\frac{{\sqrt 5 }}{3}
Option B (53\frac{{\sqrt 5 }}{3}) is correct because it is the solution to the trigonometric equation after simplification and solving for x.
127 Easy
127 of 365
Application +4 | -1
If θ=sin⁡−1x+cos⁡−1x−tan⁡−1x,  1≤x<∞\mathrm{\theta } = {\sin ^{ - 1}}x + {\cos ^{ - 1}}x - {\tan ^{ - 1}}x,\;1 \le x < \infty, then the smallest interval in which θ\mathrm{\theta } lies is
View answer and explanation
Correct option: B
0≤θ≤π40 \le \mathrm{\theta } \le \frac{\pi }{4}
Option D (π4≤θ≤π2\frac{\pi }{4} \le \mathrm{\theta } \le \frac{\pi }{2}) is incorrect. Option B (0≤θ≤π40 \le \mathrm{\theta } \le \frac{\pi }{4}) is correct because by simplifying the expression and using the given range of xx, the smallest interval in which θ\theta lies is [0,π4][0, \frac{\pi}{4}].
128 Easy
128 of 365
Application +4 | -1
The value of cot⁡−134+sin⁡−1513{\cot ^{ - 1}}\frac{3}{4} + {\sin ^{ - 1}}\frac{5}{{13}} is
View answer and explanation
Correct option: A
sin⁡−16365{\sin ^{ - 1}}\frac{{63}}{{65}}
Option A (sin⁡−16365{\sin ^{ - 1}}\frac{{63}}{{65}}) is correct because by converting both inverse trigonometric functions to inverse sine functions and applying the sine addition formula, the expression simplifies to sin⁡−16365{\sin ^{ - 1}}\frac{{63}}{{65}}.
129 Easy
129 of 365
Application +4 | -1
If x<−1/3x < -1 / \sqrt{3}, then the value of 3tan⁡−1x−tan⁡−13x−x31−3x23 \tan ^{-1} x-\tan ^{-1} \frac{3 x-x^3}{1-3 x^2} equals
View answer and explanation
Correct option: A
−π-\pi
Option A (−π-\pi) is correct because, with the given condition x<−1/3x < -1/\sqrt{3}, the expression simplifies to −π-\pi using trigonometric identities and range considerations.
130 Easy
130 of 365
Application +4 | -1
If tan⁡−12{\tan ^{ - 1}}2 and tan⁡−13{\tan ^{ - 1}}3 are two angles of a triangle, then the third angle is
View answer and explanation
Correct option: C
π4\frac{\pi }{4}
Option C (π4\frac{\pi }{4}) is correct because the sum of the angles in a triangle is π\pi, and the third angle can be found using the arctangent addition formula.
131 Easy
131 of 365
Application +4 | -1
2tan⁡−17−sin⁡−1(14/50)2 \tan ^{-1} 7-\sin ^{-1}(14 / \sqrt{50}) is equal to
View answer and explanation
Correct option: C
π\pi
Option C (π\pi) is correct because it involves using the identity for 2tan⁡−1x2\tan^{-1}x and simplifying the expression.
132 Easy
132 of 365
Analyzing +4 | -1
The number of real roots of the equation tan⁡−1x(x+1)+sin⁡−1x2+x+1=π4{\tan ^{ - 1}}\sqrt {x\left( {x + 1} \right)} \mathrm{ } + \mathrm{ }{\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = \frac{\pi }{4} is
View answer and explanation
Correct option: D
0
Option D (0) is correct because the given equation has no real roots as the domain restrictions lead to a contradiction.
133 Easy
133 of 365
Application +4 | -1
The principal value of cos⁡−1(−sin⁡7π/6)\cos ^{-1}(-\sin 7 \pi / 6) is
View answer and explanation
Correct option: C
π3\frac{\pi}{3}
Option C (π3\frac{\pi}{3}) is correct because after simplification and using the properties of inverse trigonometric functions, the principal value is found to be π3\frac{\pi}{3}.
134 Easy
134 of 365
Application +4 | -1
If sin⁡−1α+sin⁡−1β+sin⁡−1γ=3π2,  {\sin ^{ - 1}}\mathrm{\alpha } + {\sin ^{ - 1}}\mathrm{\beta } + {\sin ^{ - 1}}\mathrm{\gamma } = \frac{{3\pi }}{2},\; then αβ+αγ+βγ\mathrm{\alpha \beta } + \mathrm{\alpha \gamma } + \mathrm{\beta \gamma } is equal to
View answer and explanation
Correct option: C
33
Option A (33) is correct because when the sum of inverse sines is 3π2\frac{{3\pi }}{2}, each inverse sine must be π2\frac{\pi}{2}, leading to α=β=γ=1\alpha = \beta = \gamma = 1, and thus αβ+αγ+βγ=3\alpha\beta + \alpha\gamma + \beta\gamma = 3.
135 Easy
135 of 365
Understanding +4 | -1
The period of the function f(x)=Tan⁡−1(tan⁡x)f(x)=\operatorname{Tan}^{-1}(\tan x) is
View answer and explanation
Correct option: A
π\pi
Option A is correct because the period of tan(x) is π, and the arctangent function doesn't change the period, so the period of arctan(tan(x)) is also π.

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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
136 Easy
136 of 365
Application +4 | -1
The value of tan⁡−1(1)+cos⁡−1(−12)+sin⁡−1(−12)\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right) is equal to
View answer and explanation
Correct option: C
3π4\frac{3 \pi}{4}
Option C (3π4\frac{3 \pi}{4}) is correct because it is the simplified value of the given expression after evaluating the inverse trigonometric functions.
137 Easy
137 of 365
Analyzing +4 | -1
The solution set of the equation tan⁡−1x−cot⁡−1x=cos⁡−1(2−x)is{\tan ^{ - 1}}x - {\cot ^{ - 1}}x = {\cos ^{ - 1}}\left( {2 - x} \right)\mathrm{is}
View answer and explanation
Correct option: C
[1,3]
Option C ([1,3]) is correct because the domain of cos⁡−1(2−x){\cos^{-1}}(2-x) restricts x to the interval [1,3], and this interval satisfies the original equation.
138 Easy
138 of 365
Application +4 | -1
tan⁡−1(13)+tan⁡−1(15)+tan⁡−1(17)+tan⁡−1(18)\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right) equals
View answer and explanation
Correct option: C
π4\frac{\pi}{4}
Option C (π4\frac{{\pi}}{{4}}) is correct because applying the tangent addition formula repeatedly simplifies the expression to tan⁡−1(1)\tan^{-1}(1), which equals π4\frac{{\pi}}{{4}}.
139 Easy
139 of 365
Application +4 | -1
If −1≤x≤−12- 1 \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: C
−2  π+3cos⁡−1x- 2\;\pi + 3{\cos ^{ - 1}}x
Option C (−2;π+3cos⁡−1x- 2\\;\pi + 3{\cos ^{ - 1}}x) is correct because by substituting x=cos⁡θx = \cos \theta and considering the given range, the expression simplifies to 3θ−2π3\theta - 2\pi.
140 Easy
140 of 365
Application +4 | -1
tan⁡(2tan⁡−115+sec⁡−152+2tan⁡−118)\tan \left(2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{\sqrt{5}}{2}+2 \tan ^{-1} \frac{1}{8}\right) is equal to :
View answer and explanation
Correct option: B
2
Option B (2) is correct because by applying trigonometric identities and simplification, the expression reduces to tan(tan-1(2)) which equals 2.
141 Easy
141 of 365
Application +4 | -1
The solution of tan⁡−12x+tan⁡−13x=π4{\tan ^{ - 1}}2x + {\tan ^{ - 1}}3x = \frac{\pi }{4} is
View answer and explanation
Correct option: A
16\frac{1}{6}
Option A (16\frac{1}{6}) is correct because solving the equation tan⁡−12x+tan⁡−13x=π4{\tan ^{ - 1}}2x + {\tan ^{ - 1}}3x = \frac{\pi }{4} yields x=16x = \frac{1}{6} as the only valid solution.
142 Easy
142 of 365
Application +4 | -1
The value of cos⁡−1(cos⁡(−17π/5))\cos ^{-1}(\cos (-17 \pi / 5)), is equal to
View answer and explanation
Correct option: B
3π/53 \pi / 5
Option B (3π/53 \pi / 5) is correct because cos⁡−1(cos⁡(−17π/5))\cos^{-1}(\cos(-17\pi/5)) simplifies to cos⁡−1(cos⁡(3π/5))\cos^{-1}(\cos(3\pi/5)) which equals 3π/53\pi/5 since 3π/53\pi/5 is within the range of cos⁡−1\cos^{-1}.
143 Easy
143 of 365
Application +4 | -1
If 2sin⁡−1x=sin⁡−1(2x1−x2),{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right), then xx is equal to
View answer and explanation
Correct option: C
[−12,12]\left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]
Option C ([−12,12]\left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]) is correct because it represents the range of x for which the given equation holds true, considering the domain of the inverse sine function.
144 Easy
144 of 365
Application +4 | -1
The principal value of cos⁡−1(cos⁡2π3)+sin⁡−1(sin⁡2π3)\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right) is
View answer and explanation
Correct option: A
π\pi
Option A (π\pi) is correct because it is the result of simplifying the expression using the properties of inverse trigonometric functions.
145 Easy
145 of 365
Application +4 | -1
The value of cot⁡−1xy+1x−y+cot⁡−1yz+1y−z+cot⁡−1zx+1z−x{\cot ^{ - 1}}\frac{{xy + 1}}{{x - y}} + {\cot ^{ - 1}}\frac{{yz + 1}}{{y - z}} + {\cot ^{ - 1}}\frac{{zx + 1}}{{z - x}} is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the expression simplifies to zero due to the telescoping nature of the inverse cotangent terms.
146 Easy
146 of 365
Application +4 | -1
If tan⁡−1(13)+tan⁡−1(34)−tan⁡−1(x3)=0{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{3}{4}} \right) - {\tan ^{ - 1}}\left( {\frac{x}{3}} \right) = 0, then xx is equal to
View answer and explanation
Correct option: D
133\frac{{13}}{3}
Option D (133\frac{{13}}{3}) is correct because by applying the formula tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right), we can solve for x.
147 Easy
147 of 365
Application +4 | -1
tan⁡−134+tan⁡−135−cot⁡−18\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{3}{5}-\cot ^{-1} 8 is equal to
View answer and explanation
Correct option: A
π4\frac{\pi}{4}
Option A (π4\frac{{\pi}}{{4}}) is correct because simplifying the expression using tangent addition and subtraction formulas results in tan⁡−1(1)\tan^{-1}(1), which equals π4\frac{{\pi}}{{4}}.
148 Easy
148 of 365
Understanding +4 | -1
sec⁡−1[sec⁡(−30∘)]\sec ^{-1}\left[\sec \left(-30^{\circ}\right)\right] is equal to
View answer and explanation
Correct option: C
30∘30^{\circ}
Option C (30∘30^{\circ}) is correct because sec⁡(−30∘)=sec⁡(30∘)\sec(-30^{\circ}) = \sec(30^{\circ}) and the range of sec⁡−1x\sec^{-1}x is [0,π][0, \pi], excluding π/2\pi/2.
149 Easy
149 of 365
Application +4 | -1
If cos⁡−1x+cos⁡−1y+cos⁡−1z+cos⁡−1t=4π{\cos ^{ - 1}}x + {\cos ^{ - 1}}y + {\cos ^{ - 1}}z + {\cos ^{ - 1}}t = 4\pi, then the value of x2+y2+z2+t2{x^2} + {y^2} + {z^2} + {t^2} is
View answer and explanation
Correct option: C
4
Option C (4) is correct because the given condition cos⁡−1x+cos⁡−1y+cos⁡−1z+cos⁡−1t=4π{\cos ^{ - 1}}x + {\cos ^{ - 1}}y + {\cos ^{ - 1}}z + {\cos ^{ - 1}}t = 4\pi implies x=y=z=t=−1x = y = z = t = -1, and thus x2+y2+z2+t2=4{x^2} + {y^2} + {z^2} + {t^2} = 4.
150 Easy
150 of 365
Application +4 | -1
If −1<x<1- 1 < x < 1, then tan⁡−1(2x1−x2){\tan ^{ - 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 using the identity tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta = \frac{{2\tan \theta}}{{1 - {{\tan }^2}\theta}} and the given condition −1<x<1-1 < x < 1, we can simplify the expression.
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