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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Easy

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Showing 181-210 of 365 questions

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181 Easy
181 of 365
Analyzing +4 | -1
Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation cos⁡−1(x)−2sin⁡−1(x)=cos⁡−1(2x)\cos ^{-1}(x)-2 \sin ^{-1}(x)=\cos ^{-1}(2 x) is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because the sum of the solutions to the given equation is 0, obtained by solving the equation and summing the valid solutions.
182 Easy
182 of 365
Application +4 | -1
The value of sin⁡[2cos⁡−153]\sin \left[ {2{{\cos }^{ - 1}}\frac{{\sqrt 5 }}{3}} \right] is
View answer and explanation
Correct option: C
459\frac{{4\sqrt 5 }}{9}
Option C (459\frac{{4\sqrt 5 }}{{9}}) is correct because it is the result of simplifying the given trigonometric expression using inverse trigonometric identities.
183 Easy
183 of 365
Application +4 | -1
If cos⁡−1x+cos⁡−1y+cos⁡−1z=π,{\cos ^{ - 1}}x + {\cos ^{ - 1}}y + {\cos ^{ - 1}}z = \pi , then
View answer and explanation
Correct option: C
x2+y2+z2=1−2xyz{x^2} + {y^2} + {z^2} = 1 - 2xyz
Option D is correct because the given condition leads to the identity x2+y2+z2+2xyz=1{x^2} + {y^2} + {z^2} + 2xyz = 1, which can be rearranged to x2+y2+z2=1−2xyz{x^2} + {y^2} + {z^2} = 1 - 2xyz.
184 Easy
184 of 365
Application +4 | -1
The number of triplets (x,  y,  z)\left( {x,\;y,\;z} \right) satisfying sin⁡−1x+cos⁡−1y+sin⁡−1z=2π,  {\sin ^{ - 1}}x + {\cos ^{ - 1}}y + {\sin ^{ - 1}}z = 2\pi ,\; is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the equation is satisfied only when x = 1, y = -1, and z = 1, resulting in a single triplet.
185 Easy
185 of 365
Application +4 | -1
If −1≤x≤0- 1 \le x \le 0, then cos⁡−1(2x2−1){\cos ^{ - 1}}\left( {2{x^2} - 1} \right) equals
View answer and explanation
Correct option: C
2π−2cos⁡−1x2\pi - 2{\cos ^{ - 1}}x
Option C (2π−2cos⁡−1x2\pi - 2{\cos ^{ - 1}}x) is correct because under the given condition −1≤x≤0-1 \le x \le 0, cos⁡−1(2x2−1){\cos ^{ - 1}}(2x^2 - 1) simplifies to 2π−2cos⁡−1x2\pi - 2{\cos ^{ - 1}}x.
186 Easy
186 of 365
Application +4 | -1
If tan⁡−1x−1x+2+tan⁡−1x+1x+2=π4,{\tan ^{ - 1}}\frac{{x - 1}}{{x + 2}} + {\tan ^{ - 1}}\frac{{x + 1}}{{x + 2}} = \frac{\pi }{4}, then xx is equal to
View answer and explanation
Correct option: C
±52\pm \sqrt {\frac{5}{2}}
Option C (±52\pm \sqrt {\frac{5}{2}}) is correct because solving the given inverse tangent equation using the tangent addition formula results in x=±52x = \pm \sqrt {\frac{5}{2}}.
187 Easy
187 of 365
Application +4 | -1
sin⁡(2sin⁡−10.8)\sin \left(2 \sin ^{-1} 0.8\right) is equal to
View answer and explanation
Correct option: A
0.96
Option A (0.96) is correct because it is the result of applying the double angle formula for sine, sin⁡(2θ)=2sin⁡(θ)cos⁡(θ)\sin(2\theta) = 2\sin(\theta)\cos(\theta), with sin⁡(θ)=0.8\sin(\theta) = 0.8.
188 Easy
188 of 365
Application +4 | -1
If cos⁡−1x−cos⁡−1y2=α\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha, then 4x2−4xycos⁡α+y24 x^{2}-4 x y \cos \alpha+y^{2} is equal to
View answer and explanation
Correct option: C
4sin⁡2α4 \sin ^{2} \alpha
Option C (4sin⁡2α4 \sin ^{2} \alpha) is correct because after applying the cosine subtraction formula and simplifying, the expression equals 4sin⁡2α4 \sin^2 \alpha.
189 Easy
189 of 365
Understanding +4 | -1
sin⁡−1x+sin⁡−11x+cos⁡−1x+cos⁡−11x{\sin ^{ - 1}}x + {\sin ^{ - 1}}\frac{1}{x} + {\cos ^{ - 1}}x + {\cos ^{ - 1}}\frac{1}{x} is equal to
View answer and explanation
Correct option: A
π\pi
Option A (π\pi) is correct because using the identity sin⁡−1x+cos⁡−1x=π2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}, the expression simplifies to π2+π2=π\frac{\pi }{2} + \frac{\pi }{2} = \pi.
190 Easy
190 of 365
Application +4 | -1
If −13<x<13- \frac{1}{{\sqrt 3 }} < x < \frac{1}{{\sqrt 3 }}, then tan⁡−1(3x−x31−3x2){\tan ^{ - 1}}\left( {\frac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right) equals
View answer and explanation
Correct option: A
3tan⁡−1x3{\tan ^{ - 1}}x
Option A (3tan⁡−1x3{\tan ^{ - 1}}x) is correct because, within the given range, the expression simplifies directly using the triple angle formula for tangent.
191 Easy
191 of 365
Application +4 | -1
The value of tan⁡(2tan⁡−1(35)+sin⁡−1(513))\tan \left(2 \tan ^{-1}\left(\frac{3}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)\right) is equal to:
View answer and explanation
Correct option: A
22021\frac{220}{21}
Option A (22021\frac{{220}}{{21}}) is correct because it is the result of applying the tangent addition formula after converting the inverse sine to an inverse tangent.
192 Easy
192 of 365
Analyzing +4 | -1
If y=tan⁡−1(sec⁡x3−tan⁡x3),π2<x3<3π2,y = {\tan ^{ - 1}}(\sec {x^3} - \tan {x^3}),\frac{\pi }{2} < {x^3} < \frac{{3\pi }}{2}, then
View answer and explanation
Correct option: B
x2y′′−6y+3π2=0{x^2}y'' - 6y + \frac{{3\pi }}{2} = 0
Option B (x2y′′−6y+3π2=0{x^2}y'' - 6y + \frac{{3\pi }}{{2}} = 0) is correct because after simplifying the given equation and differentiating twice, it satisfies the given relation.
193 Easy
193 of 365
Application +4 | -1
cos⁡−1(−12)−2sin⁡−1(12)+3cos⁡−1(−12)    −4tan⁡−1(−1){\cos ^{ - 1}}\left( {\frac{{ - 1}}{2}} \right) - 2{\sin ^{ - 1}}\left( {\frac{1}{2}} \right) + 3{\cos ^{ - 1}}\left( {\frac{{ - 1}}{{\sqrt 2 }}} \right)\;\; - 4{\tan ^{ - 1}}\left( { - 1} \right) equals
View answer and explanation
Correct option: C
47π12\frac{{47\pi }}{{12}}
Option D (43π12\frac{{43\pi }}{{12}}) is correct because by evaluating each inverse trigonometric function and simplifying the expression, the result is 43π12\frac{{43\pi}}{{12}}.
194 Easy
194 of 365
Application +4 | -1
If x<0x < 0, then tan⁡−1(1x){\tan ^{ - 1}}\left( {\frac{1}{x}} \right) equals
View answer and explanation
Correct option: C
−π+cot⁡−1x- \pi + {\cot ^{ - 1}}x
Option C (−π+cot⁡−1x- \pi + {\cot ^{ - 1}}x) is correct because when x<0x < 0, the range of tan⁡−1(1x){\tan ^{ - 1}}(\frac{1}{x}) is (−π2,0)(-\frac{\pi}{2}, 0), and the correct transformation using cot⁡−1x{\cot ^{ - 1}}x accounts for this.
195 Easy
195 of 365
Application +4 | -1
If 6sin⁡−1(x2−6x+8.5)=π6 \sin ^{-1}\left(x^2-6 x+8.5\right)=\pi, then xx is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because it is one of the solutions obtained after solving the trigonometric equation and lies within the domain of the inverse sine function.

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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
196 Easy
196 of 365
Application +4 | -1
If tan⁡−1x−1x+2+tan⁡−1x+1x+2=π4,{\tan ^{ - 1}}\frac{{x - 1}}{{x + 2}} + {\tan ^{ - 1}}\frac{{x + 1}}{{x + 2}} = \frac{\pi }{4}, then xx is equal to
View answer and explanation
Correct option: C
±52\pm \sqrt {\frac{5}{2}}
Option C (±52\pm \sqrt {\frac{5}{2}}) is correct because solving the given equation using the tangent addition formula leads to this solution for x.
197 Easy
197 of 365
Application +4 | -1
If x≥1x \geq 1, then 2tan⁡−1x+sin⁡−12x1+x22 \tan ^{-1} x+\sin ^{-1} \frac{2 x}{1+x^2} is equal to
View answer and explanation
Correct option: B
π\pi
Option B (π\pi) is correct because when x ≥ 1, the expression 2tan⁻¹x + sin⁻¹(2x/(1+x²)) simplifies to π.
198 Easy
198 of 365
Application +4 | -1
5cos⁡−1(1−x21+x2)+7sin⁡−1(2x1+x2)−4tan⁡−1(2x1−x2)−tan⁡−1x=5π5{\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right) + 7{\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right) - 4{\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right) - {\tan ^{ - 1}}x = 5\pi, then xx is equal to
View answer and explanation
Correct option: D
3\sqrt 3
Option D (3\sqrt 3) is the correct answer because substituting x=3x = \sqrt{3} into the given equation satisfies the equation.
199 Easy
199 of 365
Application +4 | -1
The value of cos⁡−1(cos⁡5π3)+sin⁡−1(cos⁡5π3){\cos ^{ - 1}}\left( {\cos \frac{{5\pi }}{3}} \right) + {\sin ^{ - 1}}\left( {\cos \frac{{5\pi }}{3}} \right) is
View answer and explanation
Correct option: C
π2\frac{\pi }{2}
Option C (π2\frac{\pi}{2}) is correct because cos⁡−1x+sin⁡−1x=π2{\cos ^{ - 1}}x + {\sin ^{ - 1}}x = \frac{\pi}{2} for any xx in the domain.
200 Easy
200 of 365
Application +4 | -1
Solution of the equation cot⁡−1x+sin⁡−115=π4{\cot ^{ - 1}}x + {\sin ^{ - 1}}\frac{1}{{\sqrt 5 }} = \frac{\pi }{4} is
View answer and explanation
Correct option: A
x=3x = 3
Option A (x = 3) is correct because solving the equation using the arctangent addition formula leads to x = 3.
201 Easy
201 of 365
Application +4 | -1
Value of tan⁡−1(sin⁡2−1cos⁡2){\tan ^{ - 1}}\left( {\frac{{\sin 2 - 1}}{{\cos 2}}} \right) is
View answer and explanation
Correct option: B
1−π41 - \frac{\pi }{4}
Option B (1−π41 - \frac{\pi }{4}) is correct because by applying trigonometric identities and simplifying the expression, we find that tan⁡−1(sin⁡2−1cos⁡2)=1−π4{\tan ^{ - 1}}\left( {\frac{{\sin 2 - 1}}{{\cos 2}}} \right) = 1 - \frac{\pi }{4}.
202 Easy
202 of 365
Application +4 | -1
The value of sin⁡[π2−sin⁡−1  (−32)]\sin \left[ {\frac{\pi }{2} - {{\sin }^{ - 1}}\;\left( { - \frac{{\sqrt 3 }}{2}} \right)} \right] is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because using the complementary angle identity and simplifying gives sin⁡[π2−sin⁡−1(−32)]=12\sin \left[ {\frac{\pi}{2} - {{\sin }^{ - 1}}\left( { - \frac{{\sqrt 3 }}{2}} \right)} \right] = \frac{1}{2}.
203 Easy
203 of 365
Application +4 | -1
tan⁡[12sin⁡−1(2a1+a2)+12cos⁡−1(1−a21+a2)]  \tan \left[ {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} \right) + \frac{1}{2}{{\cos }^{ - 1}}\left( {\frac{{1 - {a^2}}}{{1 + {a^2}}}} \right)} \right]\; is equal to
View answer and explanation
Correct option: C
2a1−a2\frac{{2a}}{{1 - {a^2}}}
Option C (2a1−a2\frac{{2a}}{{1 - {a^2}}}) is correct because the expression simplifies to tan⁡(2tan⁡−1a)\tan(2{\tan ^{ - 1}}a), which equals 2a1−a2\frac{{2a}}{{1 - {a^2}}}.
204 Easy
204 of 365
Application +4 | -1
sin⁡−1(4/5)+sin⁡−1(5/13)+sin⁡−1(16/65)\sin ^{-1}(4 / 5)+\sin ^{-1}(5 / 13)+\sin ^{-1}(16 / 65) is equal to
View answer and explanation
Correct option: B
π/2\pi / 2
Option B (π/2\pi/2) is correct because the sum of the inverse sines simplifies to sin⁡−1(63/65)+sin⁡−1(16/65)\sin^{-1}(63/65) + \sin^{-1}(16/65), which equals π/2\pi/2 due to the identity sin⁡−1x+cos⁡−1x=π/2\sin^{-1}x + \cos^{-1}x = \pi/2.
205 Easy
205 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 (3π\pi/4) is correct because it is the result of evaluating the inverse trigonometric functions and summing the results.
206 Easy
206 of 365
Application +4 | -1
If sin⁡−1(1−x)−2sin⁡−1x=π2{\sin ^{ - 1}}\left( {1 - x} \right) - 2{\sin ^{ - 1}}x = \frac{\pi }{2}, then xx equals
View answer and explanation
Correct option: C
0
Option C (0) is correct because, after solving the equation, x=0x = 0 is the only solution that satisfies the original equation.
207 Easy
207 of 365
Analyzing +4 | -1
If α=sin⁡−132+sin⁡−113,  β=cos⁡−132+cos⁡−113\alpha = {\sin ^{ - 1}}\frac{{\sqrt 3 }}{2} + {\sin ^{ - 1}}\frac{1}{3},\;\beta = {\cos ^{ - 1}}\frac{{\sqrt 3 }}{2} + {\cos ^{ - 1}}\frac{1}{3}, then
View answer and explanation
Correct option: C
α<β\alpha < \beta
Option C (α<β\alpha < \beta) is correct because, given α+β=π\alpha + \beta = \pi, and by comparing the values of α\alpha and π2\frac{\pi}{2}, it can be shown that β>π2\beta > \frac{\pi}{2}, thus α<β\alpha < \beta.
208 Easy
208 of 365
Application +4 | -1
The value of cot⁡(cosec−153+tan⁡−123)\cot \left( {\mathrm{cose}{\mathrm{c}^{ - 1}}\frac{5}{3} + {{\tan }^{ - 1}}\frac{2}{3}} \right) is
View answer and explanation
Correct option: B
617\frac{6}{{17}}
Option B (617\frac{6}{{17}}) is correct because it is the simplified value of the given expression after applying trigonometric identities and the cotangent addition formula.
209 Easy
209 of 365
Application +4 | -1
The value of cos⁡−1(cos⁡5π3)+sin⁡−1(sin⁡5π3)\cos ^{-1}\left(\cos \frac{5 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{5 \pi}{3}\right) is
View answer and explanation
Correct option: A
0
Option A (0) is correct because after simplifying using trigonometric properties and the ranges of inverse trigonometric functions, the expression equals 0.
210 Easy
210 of 365
Application +4 | -1
If tan⁡−11+x2−1x=4∘\tan ^{-1} \frac{\sqrt{1+x^2}-1}{x}=4^{\circ}, then
View answer and explanation
Correct option: D
x=tan⁡8∘x=\tan 8^{\circ}
Option D (x=tan⁡8∘x=\tan 8^{\circ}) is correct because substituting x=tan⁡θx = \tan \theta and simplifying using trigonometric identities leads to θ=8∘\theta = 8^{\circ}.
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