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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Medium

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Showing 151-180 of 221 questions

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151 Medium
151 of 221
Understanding +4 | -1
Domain of f(x)=cos⁡−1x+cot⁡−1x+cosec⁡−1xf(x)=\cos ^{-1} x+\cot ^{-1} x+\operatorname{cosec}^{-1} x is
View answer and explanation
Correct option: D
{−1,1}\{-1,1\}
Option D ({−1,1-1,1}) is correct because the domain of cos⁡−1x\cos^{-1}x is [−1,1][-1, 1], the domain of cot⁡−1x\cot^{-1}x is R\mathbb{R}, and the domain of csc⁡−1x\csc^{-1}x is (−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty). The intersection of these domains is {−1,1-1, 1}.
152 Medium
152 of 221
Application +4 | -1
The value of [∣sin⁡(tan⁡−1(tan⁡3π4))∣]\left[\left|\sin \left(\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)\right)\right|\right] equals; where [.] is gint function and ∥\| is modulus function
View answer and explanation
Correct option: C
0
Option C (0) is correct because tan(3π/4) = -1, sin(tan⁻¹(-1)) = -1/√2, |sin(tan⁻¹(tan(3π/4)))| = 1/√2, and the greatest integer function [1/√2] = 0.
153 Medium
153 of 221
Application +4 | -1
If x+1/x=5/2x+1 / x=5 / 2, then the principal value of sin⁡−1x\sin ^{-1} x is
View answer and explanation
Correct option: A
π/6\pi / 6
Option A, π/6\pi / 6, is correct because by solving the quadratic equation for sin⁡x\sin x, we find that sin⁡x=1/2\sin x = 1/2, which gives x=π/6x = \pi/6.
154 Medium
154 of 221
Analyzing +4 | -1
If 0<p<2π0 < p < 2 \pi and sin⁡−1(sin⁡p)<x2y+3y2x+181xy∀x,y∈R+\sin ^{-1}(\sin p) < \frac{x^2}{y}+\frac{3 y^2}{x}+\frac{1}{81 x y} \forall x, y \in R^{+}then pp cannot be equal to
View answer and explanation
Correct option: B
1.67
Option A (0) is correct because the equation has no solution since the LHS is always non-negative while the RHS is non-positive, and they cannot both be zero simultaneously.
155 Medium
155 of 221
Application +4 | -1
If sin⁡−1(x)α=cos⁡−1xb=tan⁡−1yc,0<x<1,\frac{{{{\sin }^{ - 1}}\left( x \right)}}{\alpha } = \frac{{{{\cos }^{ - 1}}x}}{b} = \frac{{{{\tan }^{ - 1}}y}}{c},0 < x < 1, then the value of cos⁡(πca+b)\cos \left( {\frac{{\pi c}}{{a + b}}} \right) is
View answer and explanation
Correct option: C
1−y21+y2 \frac{{1 - {y^2}}}{{1 + {y^2}}}\,
Option C (1−y21+y2,\frac{{1 - {y^2}}}{{1 + {y^2}}}\\,) is correct because it is derived from trigonometric identities and the given relationships between inverse trigonometric functions.
156 Medium
156 of 221
Application +4 | -1
The value of
sin⁡−1{cot⁡(sin⁡−12−34+cos⁡−1124+sec⁡−12)}\sin ^{-1}\left\{\cot \left(\sin ^{-1} \sqrt{\frac{2-\sqrt{3}}{4}}+\cos ^{-1} \frac{\sqrt{12}}{4}+\sec ^{-1} \sqrt{2}\right)\right\} is equal to
View answer and explanation
Correct option: C
0
Option C (0) is correct because the expression simplifies to sin⁡−1(cot⁡(π/2))\sin^{-1}(\cot(\pi/2)), which equals 0.
157 Medium
157 of 221
Application +4 | -1
If x=1/5x=1 / 5, the value of cos⁡(cos⁡−1x+2sin⁡−1x)\cos \left(\cos ^{-1} x+2 \sin ^{-1} x\right) is
View answer and explanation
Correct option: C
−1/5-1 / 5
Option C (-1/5) is correct because using the identity cos⁡−1x+sin⁡−1x=π/2\cos^{-1}x + \sin^{-1}x = \pi/2, the expression simplifies to −sin⁡(sin⁡−1x)=−x=−1/5-\sin(\sin^{-1}x) = -x = -1/5.
158 Medium
158 of 221
Application +4 | -1
If cos⁡−1x+cos⁡−1y+cos⁡−1z=π\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\pi, where −1≤x,y≤1-1 \leq x, y \leq 1 and x+y≥0x+y \geq 0, then x2+y2+z2+2xyzx^2+y^2+z^2+2 x y z is equal to
View answer and explanation
Correct option: B
1
Option B, 1, is correct because by using the cosine addition formula and simplifying, we can derive that x2+y2+z2+2xyz=1x^2 + y^2 + z^2 + 2xyz = 1.
159 Medium
159 of 221
Application +4 | -1
If (tan⁡−1x)2+(cot⁡−1x)2=5π2/8\left(\tan ^{-1} x\right)^2+\left(\cot ^{-1} x\right)^2=5 \pi^2 / 8, then xx is equal to
View answer and explanation
Correct option: A
-1
The correct answer is A (-1) because solving the given equation using the relationship between tan⁡−1x\tan^{-1} x and cot⁡−1x\cot^{-1} x leads to x = -1.
160 Medium
160 of 221
Analyzing +4 | -1
If
tan⁡−1(x+h)=tan⁡−1(x)+(hsin⁡y)(sin⁡y)−(hsin⁡y)2⋅sin⁡2y2+(hsin⁡y)3⋅sin⁡3y3+………\tan ^{-1}(x+h)=\tan ^{-1}(x)+(h \sin y)(\sin y)-(h \sin y)^2 \cdot \frac{\sin 2 y}{2}+(h \sin y)^3 \cdot \frac{\sin 3 y}{3}+\ldots \ldots \ldots

Where x∈(0,1),y∈(π/4,π/2)x \in(0,1), y \in(\pi / 4, \pi / 2), then
View answer and explanation
Correct option: C
y=cot⁡−1xy=\cot ^{-1} x
Option C (y=cot⁡−1xy=\cot^{-1} x) is correct because by comparing the given series with the Taylor series expansion of tan⁡−1(x+h)\tan^{-1}(x+h), we can deduce that y=cot⁡−1xy = \cot^{-1}x.
161 Medium
161 of 221
Application +4 | -1
If 0<x<10 < x < 1, then 1+x2[{xcos⁡(cot⁡−1x)\sqrt{1+x^2}\left[\left\{x \cos \left(\cot ^{-1} x\right)\right.\right. +sin⁡(cot⁡−1x)}2−1]1/2\left.\left.+\sin \left(\cot ^{-1} x\right)\right\}^2-1\right]^{1 / 2} is equal to
View answer and explanation
Correct option: C
x1+x2x \sqrt{1+x^2}
Option C (x1+x2x \sqrt{1+x^2}) is correct because by substituting cot-1(x) with theta and simplifying the expression using trigonometric identities, the final result is x*sqrt(1+x2).
162 Medium
162 of 221
Application +4 | -1
Number of solutions of the equation
tan⁡−1(12x+1)+tan⁡−1(14x+1)=tan⁡−1(2x2){\tan ^{ - 1}}\left( {\frac{1}{{2x + 1}}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{{4x + 1}}} \right) = {\tan ^{ - 1}}\left( {\frac{2}{{{x^2}}}} \right) is
View answer and explanation
Correct option: B
2
The correct answer is 2 because solving the equation leads to three possible solutions, but one of them is extraneous and does not satisfy the original equation.
163 Medium
163 of 221
Analyzing +4 | -1
If 0<a,  b<10 < a,\,\,b < 1 and tan⁡−1a+tan⁡−1b=π4{\tan ^{ - 1}}a + {\tan ^{ - 1}}b = \frac{\pi }{4}, then the value of (a+b)−(a2+b22)+(a3+b33)−(a4+b44)+ ...... \left( {a + b} \right) - \left( {\frac{{{a^2} + {b^2}}}{2}} \right) + \left( {\frac{{{a^3} + {b^3}}}{3}} \right) - \left( {\frac{{{a^4} + {b^4}}}{4}} \right) + \,......\,. is
View answer and explanation
Correct option: A
log⁡e2{\log _e}2
Option A (log⁡e2{\log _e}2) is correct because the given condition leads to (1+a)(1+b) = 2, and the series can be expressed as log(1+a) + log(1+b) = log(2).
164 Medium
164 of 221
Application +4 | -1
Let a∈(−π2,π2)a \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right) such that tan⁡−1(tan⁡α3+2tan⁡2α)+tan⁡−1(2tan⁡α3)=π12\tan ^{-1}\left(\frac{\tan \alpha}{3+2 \tan ^2 \alpha}\right)+\tan ^{-1}\left(\frac{2 \tan \alpha}{3}\right)=\frac{\pi}{12}, then α\alpha equals:
View answer and explanation
Correct option: D
π12\frac{\pi}{12}
Option D (π12\frac{\pi}{12}) is correct because by simplifying the given equation using trigonometric identities, we find that α\alpha equals π12\frac{\pi}{12}.
165 Medium
165 of 221
Analyzing +4 | -1
The number of real solution of cot⁡−1x(x+4)+cos⁡−1x2+4x+1=π2\cot ^{-1} \sqrt{x(x+4)}+\cos ^{-1} \sqrt{x^2+4 x+1}=\frac{\pi}{2} is equal to
View answer and explanation
Correct option: C
2
Option C (2) is correct because the equation is satisfied only when x2+4x=0x^2 + 4x = 0, leading to two real solutions: 0 and -4.

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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
166 Medium
166 of 221
Application +4 | -1
If −12≤x≤12- \frac{1}{2} \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: B
2π−3cos⁡−1x2\pi - 3{\cos ^{ - 1}}x
Option D (None of these) is correct because the correct expression is 3cos⁡−1x3{\cos ^{ - 1}}x when 0≤x≤10 \leq x \leq 1, and 2π−3cos⁡−1x2\pi - 3{\cos ^{ - 1}}x when −1≤x≤−1/2-1 \leq x \leq -1/2. Since the given range is −1/2≤x≤1/2-1/2 \le x \le 1/2, the expression is 3cos⁡−1x3\cos^{-1}x only when 1/2≤x≤11/2 \le x \le 1.
167 Medium
167 of 221
Application +4 | -1
The xx satisfying sin⁡−1x+sin⁡−1(1−x)=cos⁡−1x\sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x are
View answer and explanation
Correct option: C
0,120, \frac{1}{2}
Option C (0,120, \frac{{1}}{{2}}) is correct because these values satisfy the given equation when substituted and verified using inverse trigonometric properties.
168 Medium
168 of 221
Application +4 | -1
If x<0x < 0, then tan⁡−1x+tan⁡−1(1/x)\tan ^{-1} x+\tan ^{-1}(1 / x) equals
View answer and explanation
Correct option: C
−π/2-\pi / 2
The correct answer is C (−π/2-\pi / 2) because when x < 0, the identity tan⁡−1x+tan⁡−1(1/x)=−π/2\tan^{-1} x + \tan^{-1}(1/x) = -\pi/2 holds.
169 Medium
169 of 221
Application +4 | -1
If 12sin⁡−1[3sin⁡2θ5+4cos⁡2θ]=tan⁡−1x\frac{1}{2} \sin ^{-1}\left[\frac{3 \sin 2 \theta}{5+4 \cos 2 \theta}\right]=\tan ^{-1} x, then x=x=
View answer and explanation
Correct option: C
(1/3)tan⁡θ(1 / 3) \tan \theta
Option C ((1/3)tan θ) is correct because by simplifying the expression inside the inverse sine function, we can find the value of x.
170 Medium
170 of 221
Application +4 | -1
If sin⁡−1(sin⁡x)+cos⁡−1(cos⁡y)+sec⁡−1(sec⁡z)=5π2\sin ^{-1}(\sin x)+\cos ^{-1}(\cos y)+\sec ^{-1}(\sec z)=\frac{5 \pi}{2} then the value of sin⁡x+cos⁡y+sec⁡z\sin x+\cos y+\sec z is
View answer and explanation
Correct option: B
-1
Option B (-1) is correct because the given equation is only possible if sin x = 1, cos y = -1, and sec z = -1, leading to sin x + cos y + sec z = -1.
171 Medium
171 of 221
Application +4 | -1
If sin⁡−12a1+a2+sin⁡−12b1+b2=2tan⁡−1x,\sin ^{-1} \frac{2 a}{1+a^2}+\sin ^{-1} \frac{2 b}{1+b^2}=2 \tan ^{-1} x, then xx is equal to
View answer and explanation
Correct option: D
a+b1−ab\frac{a+b}{1-a b}
The correct answer is B (x+2y+4=0x+2 y+4=0) because it represents a common tangent to both the parabola and the ellipse.
172 Medium
172 of 221
Application +4 | -1
The number of solutions of equation 2(sin⁡−1x)2−5sin⁡−1x+2=02\left(\sin ^{-1} x\right)^2-5 \sin ^{-1} x+2=0 is
View answer and explanation
Correct option: B
1
Option B (1) is correct because only one solution for xx exists within the domain of sin⁡−1x\sin^{-1} x, which is [−1,1][-1, 1].
173 Medium
173 of 221
Application +4 | -1
If we consider only the principle value of the inverse trigonometric functions, then the value of tan⁡(cos⁡−1152−sin⁡−1417)\tan \left( {{{\cos }^{ - 1}}\frac{1}{{5\sqrt 2 }} - {{\sin }^{ - 1}}\frac{4}{{\sqrt {17} }}} \right) is
View answer and explanation
Correct option: D
329\frac{3}{{29}}
Option D (329\frac{3}{{29}}) is correct because the expression simplifies using trigonometric identities and the tangent subtraction formula.
174 Medium
174 of 221
Analyzing +4 | -1
The domain of the function Cos⁡−1(∣x∣−32)+1log⁡10(4−x)\operatorname{Cos}^{-1}\left(\frac{|x|-3}{2}\right)+\frac{1}{\log _{10}(4-x)} is
View answer and explanation
Correct option: A
[−5,−1]∪[1,3)∪(3,4)[-5,-1] \cup[1,3) \cup(3,4)
Option A ([-5,-1] ∪ [1,3) ∪ (3,4)) is correct because it represents the intersection of the domains of the inverse cosine and the logarithmic functions.
175 Medium
175 of 221
Application +4 | -1
The value of tan⁡{cos⁡−145+tan⁡−125}\tan \left\{\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{5}\right\} is λμ;(λ,μ)=1\frac{\lambda}{\mu} ;(\lambda, \mu)=1, then evaluate λ+19μ\frac{\lambda+19}{\mu}.
View answer and explanation
Correct option: A
3
Option A (3) is correct because the expression simplifies to 2314\frac{23}{14}, so λ=23\lambda = 23 and μ=14\mu = 14, and λ+19μ=3\frac{\lambda+19}{\mu} = 3.
176 Medium
176 of 221
Analyzing +4 | -1
Let f(x)=sin⁡−1(sin⁡x)f(x)=\sin ^{-1}(\sin x) and g(x)=cos⁡−1[x+1]+sin⁡−1[x]g(x)=\cos ^{-1}[x+1]+\sin ^{-1}[x]. Then the number of solution of f(x)+g(x)=4f(x)+g(x)=4, lies in the interval

(where [∗]\left[{ }^*\right] denotes the greatest integer function)
View answer and explanation
Correct option: C
(−2,1)(-2,1)
Option C, (−2,1)(-2,1), is correct because the domain of g(x) is [-1, 1) and within this domain, g(x) = 0, leading to no solution for f(x) + g(x) = 4.
177 Medium
177 of 221
Application +4 | -1
If tan⁡−1n,tan⁡−1(n+1)\tan ^{-1} \mathrm{n}, \tan ^{-1}(\mathrm{n}+1) and tan⁡−1(n+2),n∈N\tan ^{-1}(n+2), \mathrm{n} \in \mathbb{N}, are angles of a triangle, then find n\mathrm{n}.
View answer and explanation
Correct option: A
1
Option A (1) is correct because when n=1, the angles are tan-1(1), tan-1(2), and tan-1(3), which can be angles of a triangle.
178 Medium
178 of 221
Application +4 | -1
The value of cos⁡−1(−1/2)−2sin⁡−1(1/2)+\cos ^{-1}(-1 / 2)-2 \sin ^{-1}(1 / 2)+ 3cos⁡−1(−1/2)−4tan⁡−1(−1)3 \cos ^{-1}(-1 / \sqrt{2})-4 \tan ^{-1}(-1) is equal to
View answer and explanation
Correct option: D
25π/1225 \pi / 12
Option D (25π/1225 \pi / 12) is correct because the expression evaluates to (2π/3)−2×(π/6)+3×(3π/4)−4(−π/4)=25π/12(2 \pi / 3)-2 \times (\pi / 6)+3 \times (3 \pi / 4)-4(-\pi / 4)=25 \pi/12.
179 Medium
179 of 221
Application +4 | -1
Let (sin⁡a)x2+(sin⁡a)x+(1−cos⁡a)=0(\sin a) x^2+(\sin a) x+(1-\cos a)=0. The value of a for which roots of this equation are real and distinct, is
View answer and explanation
Correct option: A
(0,2tan⁡−11/4]\left(0,2 \tan ^{-1} 1 / 4\right]
Option A ((0,2tan⁡−11/4]\left(0,2 \tan ^{{-1}} 1 / 4\right]) is correct because it represents the range of 'a' for which the discriminant of the quadratic equation is positive, ensuring real and distinct roots.
180 Medium
180 of 221
Analyzing +4 | -1
Let I1=π24+2,  I2=(tan⁡−1(1e))2+2ee2+1,I3=(tan⁡−1e)2+2e2+1I_1=\frac{\pi^2}{4}+\sqrt{2}, \, \, I_2=\left(\tan ^{-1}\left(\frac{1}{e}\right)\right)^2+\frac{2 e}{\sqrt{e^2+1}}, I_3=\left(\tan ^{-1} e\right)^2+\frac{2}{\sqrt{e^2+1}}, then which of the following is true?
View answer and explanation
Correct option: B
I2<I1<I3{I}_2 < {I}_1 < {I}_3
Option B (I2<I1<I3{I}_2 < {I}_1 < {I}_3) is correct because by analyzing the function f(x)=(tan⁡−1x)2+2x2+1f(x)=\left(\tan ^{-1} x\right)^2+\frac{2}{\sqrt{x^2+1}} and its derivative, we can determine the correct inequality.
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