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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Mixed difficulty

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121 Medium
121 of 605
Application +4 | -1
If (sin⁡−1x)2+(sin⁡−1y)2+2sin⁡−1xsin⁡−1y=π2\left(\sin ^{-1} x\right)^2+\left(\sin ^{-1} y\right)^2+2 \sin ^{-1} x \sin ^{-1} y=\pi^2, then x2+y2x^2+y^2 is equal to:
View answer and explanation
Correct option: C
2
Option C (2) is correct because the given equation can be simplified to find the values of x and y, and then x2+y2x^2 + y^2 can be calculated.
122 Medium
122 of 605
Application +4 | -1
The number of positive integral pairs (a,b)(a, b) satisfying the equation tan⁡−1a+tan⁡−1b=tan⁡−17\tan ^{-1} a+\tan ^{-1} b=\tan ^{-1} 7 is
View answer and explanation
Correct option: B
2
Option B, 2, is correct because by applying the tangent addition formula and solving for a in terms of b, we find only two positive integer pairs (a, b) that satisfy the equation.
123 Easy
123 of 605
Analyzing +4 | -1
cot⁡−1(2.12)+cot⁡−1(2.22)+cot⁡−1(2.32)+...{\cot ^{ - 1}}\left( {{{2.1}^2}} \right) + {\cot ^{ - 1}}\left( {{{2.2}^2}} \right) + {\cot ^{ - 1}}\left( {{{2.3}^2}} \right) + ... upto ∞\infty is equal to
View answer and explanation
Correct option: A
π4\frac{\pi }{4}
Option A (π4\frac{\pi }{4}) is correct because the series can be expressed as a telescoping sum of arctangents, which simplifies to π4\frac{\pi }{4}.
124 Hard
124 of 605
Application +4 | -1
If f(x)=sin⁡−1{32x−121−x2},−72≤x≤1f(x)=\sin ^{-1}\left\{\frac{\sqrt{3}}{2} x-\frac{1}{2} \sqrt{1-x^2}\right\},-\frac{7}{2} \leq x \leq 1, then f(x)f(x) is equal to f(x)f(x) is equal to
View answer and explanation
Correct option: B
sin⁡−1x−π6\sin ^{-1} x-\frac{\pi}{6}
Option B (sin⁡−1x−π6\sin ^{-1} x-\frac{\pi}{6}) is correct because the given expression simplifies to sin⁡−1x−π6\sin^{-1} x - \frac{\pi}{6} using trigonometric identities and substitution.
125 Easy
125 of 605
Analyzing +4 | -1
If a,  b,  ca,\;b,\;c be positive real number and the value of

θ=tan⁡−1a(a+b+c)bc+tan⁡−1b(a+b+c)ca+tan⁡−1c(c+b+c)ab\mathrm{\theta } = {\tan ^{ - 1}}\sqrt {\frac{{a\left( {a + b + c} \right)}}{{bc}}} + {\tan ^{ - 1}}\sqrt {\frac{{b\left( {a + b + c} \right)}}{{ca}}} + {\tan ^{ - 1}}\sqrt {\frac{{c\left( {c + b + c} \right)}}{{ab}}}

Then tan θ\mathrm{\theta } is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because after substituting and simplifying, the expression inside the tangent function becomes zero, resulting in tan θ = 0.
126 Easy
126 of 605
Application +4 | -1
If −π2<sin⁡−1x<π2- \frac{\pi }{2} < {\sin ^{ - 1}}x < \frac{\pi }{2}, then tan⁡(sin⁡−1x)\tan \left( {{{\sin }^{ - 1}}x} \right) is equal to
View answer and explanation
Correct option: C
x1−x2\frac{x}{{\sqrt {1 - {x^2}} }}
Option C (x1−x2\frac{x}{{\sqrt {1 - {x^2}} }}) is correct because it represents the tangent of the angle whose sine is x, simplified using trigonometric identities.
127 Easy
127 of 605
Application +4 | -1
If α,  β\alpha ,\;\beta are the roots of the equation 6x2−5x+1=06{x^2} - 5x + 1 = 0, then the value of tan⁡−1α+tan⁡−1β{\tan ^{ - 1}}\alpha + {\tan ^{ - 1}}\beta is
View answer and explanation
Correct option: B
π/4\pi /4
Option B (π/4\pi /4) is correct because by finding the sum and product of the roots and using the formula for tan⁡−1α+tan⁡−1β{\tan ^{ - 1}}\alpha + {\tan ^{ - 1}}\beta, we can simplify the expression to tan⁡−1(1){\tan ^{ - 1}}(1), which equals π4\frac{\pi}{4}.
128 Easy
128 of 605
Application +4 | -1
If x≥1x \ge 1, then 2tan⁡−1x+sin⁡−1(2x1+x2)2{\tan ^{ - 1}}x + {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right) is equal to
View answer and explanation
Correct option: D
π\pi
Option D is correct because for x≥1x \ge 1, sin⁡−1(2x1+x2)=π−2tan⁡−1x{\sin ^{ - 1}}(\frac{{2x}}{{1 + {x^2}}}) = \pi - 2{\tan ^{ - 1}}x, and thus the expression simplifies to π\pi.
129 Easy
129 of 605
Remembering +4 | -1
Prinicpal value of tan⁡−1(3)\tan ^{-1}(\sqrt{3}) is equal to
View answer and explanation
Correct option: B
π3\frac{\pi}{3}
Option B (π3\frac{\pi}{3}) is correct because the principal value of tan⁡−1(3)\tan^{-1}(\sqrt{3}) lies in the range (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}) and tan⁡(π3)=3\tan(\frac{\pi}{3}) = \sqrt{3}.
130 Medium
130 of 605
Application +4 | -1
The value of sin⁡−1[cos⁡(tan⁡−113+tan⁡−112)]\sin ^{-1}\left[\cos \left(\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{2}\right)\right] is π/k\pi / k; then find the value of kk.
View answer and explanation
Correct option: A
4
Option A (4) is correct because the expression simplifies to sin⁡−1(cos⁡(π4))=sin⁡−1(22)=π4\sin^{-1}(\cos(\frac{\pi}{4})) = \sin^{-1}(\frac{\sqrt{2}}{2}) = \frac{\pi}{4}, thus k=4.
131 Easy
131 of 605
Analyzing +4 | -1
The number of real solution of tan⁡−1x(x+1)+sin⁡−1  x2+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 is satisfied for two values of x, namely x = 0 and x = -1.
132 Hard
132 of 605
Application +4 | -1
The value tan⁡(12cos⁡−153)\tan \left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right) is
View answer and explanation
Correct option: B
3−52\frac{3-\sqrt{5}}{2}
Option B ((3-√5)/2) is correct because using the half-angle formula for tangent and simplifying leads to the correct result.
133 Medium
133 of 605
Application +4 | -1
If tan⁡−12p+tan⁡−13p+tan⁡−14p+tan⁡−15p=π2\tan ^{-1} \frac{2}{p}+\tan ^{-1} \frac{3}{p}+\tan ^{-1} \frac{4}{p}+\tan ^{-1} \frac{5}{p}=\frac{\pi}{2}, then (p2+120p2)\left(p^2+\frac{120}{p^2}\right) is equal to
View answer and explanation
Correct option: B
71
Option B (71) is correct because solving the given equation involving inverse tangents leads to the equation p4+120=71p2p^4 + 120 = 71p^2, which directly gives p2+120p2=71p^2 + \frac{120}{p^2} = 71.
134 Easy
134 of 605
Application +4 | -1
The value of cos⁡[tan⁡−1{sin⁡(cot⁡−1x)}]\cos \left[ {{{\tan }^{ - 1}}\left\{ {\sin \left( {{{\cot }^{ - 1}}x} \right)} \right\}} \right] is
View answer and explanation
Correct option: D
x2+1x2+2\sqrt {\frac{{{x^2} + 1}}{{{x^2} + 2}}}
Option D is correct because simplifying the expression using trigonometric identities leads to x2+1x2+2\sqrt {\frac{{{x^2} + 1}}{{{x^2} + 2}}}.
135 Medium
135 of 605
Application +4 | -1
Iftan⁡−111+2+tan⁡−111+(2)(3)+tan⁡−111+(3)(4)+…+tan⁡−111+n(n+1)=tan⁡−1θ then θ=\text{If} \tan ^{-1} \frac{1}{1+2}+\tan ^{-1} \frac{1}{1+(2)(3)}+\tan ^{-1} \frac{1}{1+(3)(4)}+\ldots+\tan ^{-1} \frac{1}{1+n(n+1)}=\tan ^{-1}\theta ~\text{then} ~\theta=
View answer and explanation
Correct option: C
nn+2\frac{n}{n+2}
Option C (nn+2\frac{n}{n+2}) is correct because the given sum telescopes, simplifying to an expression whose arctangent is easily calculated.

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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 605
Analyzing +4 | -1
Which one of the following is correct?
View answer and explanation
Correct option: A
tan⁡1>tan⁡−11\tan 1 > {\tan ^{ - 1}}1
Option A (tan⁡1>tan⁡−11\tan 1 > {\tan ^{ - 1}}1) is correct because tan⁡1\tan 1 (where 1 is in radians) is greater than 1, while tan⁡−11=π4\tan^{-1} 1 = \frac{\pi}{4} which is less than 1.
137 Hard
137 of 605
Application +4 | -1
If tan⁡−1x+cos⁡−1y(1+y2)=sin⁡−1310\tan ^{-1} x+\cos ^{-1} \frac{y}{\sqrt{\left(1+y^2\right)}}=\sin ^{-1} \frac{3}{\sqrt{10}} and both xx and yy are positive and integral, then xx and yy is equal to
View answer and explanation
Correct option: A
(1,2)(1,2) and (2,7)(2,7)
Option A ((1,2) and (2,7)) is correct because solving the given equation with the constraints that x and y are positive integers leads to these two solution pairs.
138 Easy
138 of 605
Application +4 | -1
If sin⁡−1(x5)+cosec−1(54)=π2{\sin ^{ - 1}}\left( {\frac{x}{5}} \right) + \mathrm{cose}{\mathrm{c}^{ - 1}}\left( {\frac{5}{4}} \right) = \frac{\pi }{2}, then value of xx is
View answer and explanation
Correct option: B
3
Option B (3) is correct because by using the property that sin⁡−1x+csc⁡−1y=π2{\sin ^{ - 1}}x + {\csc ^{ - 1}}y = \frac{\pi }{2} implies x=1yx = \frac{1}{y}, we can solve for x.
139 Medium
139 of 605
Application +4 | -1
The value of cot⁡[∑n=123cot⁡−1(1+∑k=1n2k)]\cot \left[\sum\limits_{{n=1}}^{23} \cot ^{-1}\left(1+\sum\limits_{{k=1}}^n 2 k\right)\right] is
View answer and explanation
Correct option: B
2523\frac{25}{23}
Option B (25/23) is correct because the summation of the inverse cotangent series simplifies to a single inverse cotangent, and taking the cotangent of that result yields 25/23.
140 Easy
140 of 605
Application +4 | -1
Let a1=1,a2,a3,a4,…a_1=1, a_2, a_3, a_4, \ldots be consecutive natural numbers. Then

tan⁡−1(11+a1a2)+tan⁡−1(11+a2a3)+…+tan⁡−1(11+a2021a2022) is equal to \tan ^{-1}\left(\frac{1}{1+a_1 a_2}\right) +\tan ^{-1}\left(\frac{1}{1+a_2 a_3}\right)+\ldots +\tan ^{-1}\left(\frac{1}{1+a_{2021} a_{2022}}\right) \text { is equal to }
View answer and explanation
Correct option: A
tan⁡−1(2022)−π4\tan ^{-1}(2022)-\frac{\pi}{4}
Option A (tan⁡−1(2022)−π4\tan ^{-1}(2022)-\frac{{\pi}}{{4}}) is correct because the given sum telescopes to tan⁡−1(a2022)−tan⁡−1(a1)=tan⁡−1(2022)−tan⁡−1(1)=tan⁡−1(2022)−π4\tan^{-1}(a_{2022}) - \tan^{-1}(a_1) = \tan^{-1}(2022) - \tan^{-1}(1) = \tan^{-1}(2022) - \frac{{\pi}}{{4}}.
141 Easy
141 of 605
Application +4 | -1
If −1≤x≤−12- 1 \le x \le - \frac{1}{{\sqrt 2 }}, then sin⁡−1(2x1−x2){\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right) equals
View answer and explanation
Correct option: C
−π−2sin⁡−1x- \pi - 2{\sin ^{ - 1}}x
Option C (−π−2sin⁡−1x- \pi - 2{\sin ^{ - 1}}x) is correct because, within the given range for x, the expression sin⁡−1(2x1−x2){\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} ) simplifies to this value after considering the range of the inverse sine function.
142 Easy
142 of 605
Application +4 | -1
If cos⁡−1(23x)+cos⁡−1(34x)=π2(x>34)\cos ^{-1}\left(\frac{2}{3 x}\right)+\cos ^{-1}\left(\frac{3}{4 x}\right)=\frac{\pi}{2}\left(x>\frac{3}{4}\right), then xx is equal to:
View answer and explanation
Correct option: D
14512\frac{\sqrt{145}}{12}
Option D (14512\frac{\sqrt{{145}}}{{12}}) is correct because solving the given equation using trigonometric identities leads to x=14512x = \frac{\sqrt{{145}}}{{12}}.
143 Medium
143 of 605
Application +4 | -1
tan⁡(π4+12cos⁡−1x)+tan⁡(π4−12cos⁡−1x),x≠0\tan \left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x\right)+\tan \left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\right), x \neq 0, is equal to
View answer and explanation
Correct option: C
2x\frac{2}{x}
Option C (2x\frac{2}{x}) is correct because the given expression simplifies to 2x\frac{2}{x} using trigonometric identities and substitution.
144 Easy
144 of 605
Application +4 | -1
The relation tan⁡−1(1+x1−x)=π4+tan⁡−1x{\tan ^{ - 1}}\left( {\frac{{1 + x}}{{1 - x}}} \right) = \frac{\pi }{4} + {\tan ^{ - 1}}x holds true for all
View answer and explanation
Correct option: B
x∈(−∞,  1)x \in \left( { - \infty ,\;1} \right)
Option B (x∈(−∞,  1)x \in \left( { - \infty ,\;1} \right)) is correct because the identity holds true only when x<1x < 1 due to the domain restrictions of the inverse tangent function.
145 Easy
145 of 605
Analyzing +4 | -1
If x>13x > \frac{1}{{\sqrt 3 }}, then x>13x > \frac{1}{{\sqrt 3 }} equals
View answer and explanation
Correct option: B
−π+3tan⁡−1x- \pi + 3{\tan ^{ - 1}}x
Option B (−π+3tan⁡−1x- \pi + 3{\tan ^{ - 1}}x) is correct because it represents the correct simplification of the given expression when x>13x > \frac{1}{{\sqrt 3 }}.
146 Easy
146 of 605
Application +4 | -1
If cos⁡−1x>sin⁡−1x,  then{\cos ^{ - 1}}x > {\sin ^{ - 1}}x,\;\mathrm{then}
View answer and explanation
Correct option: D
−1≤x<12- 1 \le x < \frac{1}{{\sqrt 2 }}
Option D (−1≤x<12- 1 \le x < \frac{1}{{\sqrt 2 }}) is correct because it represents the range of x where the inverse cosine is greater than the inverse sine.
147 Easy
147 of 605
Application +4 | -1
If 3+i=(a+ib)(c+id)\sqrt 3 + i = \left( {a + ib} \right)\left( {c + id} \right), then tan⁡−1(ba)+tan⁡−1(dc){\tan ^{ - 1}}\left( {\frac{b}{a}} \right) + {\tan ^{ - 1}}\left( {\frac{d}{c}} \right) has the value
View answer and explanation
Correct option: B
nπ+π6,  n∈In\pi + \frac{\pi }{6},\;n \in I
Option B (nπ+π6,  n∈In\pi + \frac{\pi }{6},\;n \in I) is correct because it is derived by using the tangent addition formula and simplifying the given expression.
148 Easy
148 of 605
Application +4 | -1
Let x∗y=x2+y3{x^*}y = {x^2} + {y^3} and (x∗1)∗1=x∗(1∗1).{({x^*}1)^*}1 = {x^*}({1^*}1). Then a value of 2sin⁡−1(x4+x2−2x4+x2+2)2{\sin ^{ - 1}}\left( {\frac{{{x^4} + {x^2} - 2}}{{{x^4} + {x^2} + 2}}} \right) is
View answer and explanation
Correct option: B
π3\frac{\pi }{3}
Option B (π3\frac{\pi}{3}) is correct because by substituting the given operation into the equation and solving for x, we can find the value of the expression.
149 Medium
149 of 605
Application +4 | -1
The value of sin⁡−1(sin⁡10)\sin ^{-1}(\sin 10) is
View answer and explanation
Correct option: B
3π−103 \pi-10
Option B (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 10 is outside this range. Therefore, we need to find an equivalent angle within this range, which is 3π−103\pi - 10.
150 Easy
150 of 605
Application +4 | -1
If cosec⁡−1x=2cot⁡−17+cos⁡−1\operatorname{cosec}^{-1} x=2 \cot ^{-1} 7+\cos ^{-1} (3/5) then the value of xx is
View answer and explanation
Correct option: B
125/117125 / 117
Option B (125/117) is correct because, by applying the cotangent addition formula and simplifying, we find that cosec-1x = cosec-1(125/117), thus x = 125/117.
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