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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Medium

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Showing 61-90 of 221 questions
61 Medium
61 of 221
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.
62 Medium
62 of 221
Application +4 | -1
If sin⁡−1x+sin⁡−1y=π\sin ^{-1} x+\sin ^{-1} y=\pi and, if x=λyx=\lambda y, then the value of 392λ+5239^{2 \lambda}+5^2 must be:
View answer and explanation
Correct option: B
1546
Option B (1546) is correct because from the given condition, we find that x=yx=y, so λ=1\lambda = 1, and then calculate 392λ+52=392+25=1521+25=154639^{2\lambda} + 5^2 = 39^2 + 25 = 1521 + 25 = 1546.
63 Medium
63 of 221
Application +4 | -1
tan⁡−1[cos⁡−1(45)+sin⁡−1(23)]\tan ^{-1}\left[\cos ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{2}{\sqrt{3}}\right)\right] is
View answer and explanation
Correct option: A
176\frac{17}{6}
Option D (None of these) is incorrect; the correct answer is 17/6, which is obtained by simplifying the expression using trigonometric identities.
64 Medium
64 of 221
Application +4 | -1
The principal value of cos⁡−1{12[cos⁡(9π10)−sin⁡(9π10)]}\cos ^{-1}\left\{\frac{1}{\sqrt{2}}\left[\cos \left(\frac{9 \pi}{10}\right)-\sin \left(\frac{9 \pi}{10}\right)\right]\right\} is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the principal value of the given expression simplifies to 17π20\frac{{17\pi}}{{20}}, which is not among the provided options.
65 Medium
65 of 221
Application +4 | -1
If cos⁡−1pa+cos⁡−1qb=α\cos ^{-1} \frac{p}{a}+\cos ^{-1} \frac{q}{b}=\alpha and p2a2−2pqabcos⁡α+q2 b2=f(α)\frac{\mathrm{p}^2}{\mathrm{a}^2}-\frac{2 \mathrm{pq}}{\mathrm{ab}} \cos \alpha+\frac{\mathrm{q}^2}{\mathrm{~b}^2}=\mathrm{f}(\alpha), then number of integers in the range of f(α)f(\alpha) is
View answer and explanation
Correct option: A
2
Option A (1) is correct because by simplifying the integral using trigonometric substitutions, we find that A = 1/2 and B = 1/2, so A+B = 1.
66 Medium
66 of 221
Analyzing +4 | -1
If tan⁡−1∣a∣=tan⁡−1(a2+k)−π4\tan ^{-1}|a|=\tan ^{-1}\left(a^2+k\right)-\frac{\pi}{4} holds for integral values of a∈[−2,2]a \in[-2,2]. Then number of integral value(s) of kk is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because only one integral value of k satisfies the given equation for the specified range of a.
67 Medium
67 of 221
Analyzing +4 | -1
If α=sin⁡−145+sin⁡−113and  β=cos⁡−145+cos⁡−113,  \mathrm{\alpha } = {\sin ^{ - 1}}\frac{4}{5} + {\sin ^{ - 1}}\frac{1}{3}\mathrm{and}\;\mathrm{\beta } = {\cos ^{ - 1}}\frac{4}{5} + {\cos ^{ - 1}}\frac{1}{3},\; then
View answer and explanation
Correct option: A
α<β\mathrm{\alpha } < \beta
Option A is correct because by calculating alpha and beta and comparing them, it is found that alpha < beta.
68 Medium
68 of 221
Application +4 | -1
The principal value of sin⁡−1(−3/2)+cos⁡−1cos⁡(7π/6)\sin ^{-1}(-\sqrt{3} / 2)+\cos ^{-1} \cos (7 \pi / 6) is
View answer and explanation
Correct option: B
π/2\pi / 2
Option B (π/2\pi / 2) is correct because sin⁡−1(−3/2)=−π/3\sin^{-1}(-\sqrt{3} / 2) = -\pi / 3 and cos⁡−1(cos⁡(7π/6))=5π/6\cos^{-1}(\cos(7\pi / 6)) = 5\pi / 6, and their sum is π/2\pi / 2.
69 Medium
69 of 221
Analyzing +4 | -1
Statement - I If x=sin⁡−1(sin⁡10)x=\sin ^{-1}(\sin 10) and y=cos⁡−1(cos⁡10)y=\cos ^{-1}(\cos 10) then y−x=πy-x=\pi

Statement - II The no. of solutions of the equation sin⁡−1[x2+13]+cos⁡−1[x2−23]=x2,x∈[−1,1]\sin ^{-1}\left[x^2+\frac{1}{3}\right]+\cos ^{-1}\left[x^2-\frac{2}{3}\right]=x^2, x \in[-1,1] and [.] denoted the greatest integer less than and equal to x\mathrm{x} is 0
View answer and explanation
Correct option: C
Both statements are true
Option C (Both statements are true) is correct because both the simplification of inverse trigonometric functions and the analysis of the equation involving greatest integer functions lead to valid conclusions.
70 Medium
70 of 221
Application +4 | -1
If sec⁡−11+x2+cosec−11+y2y+cot⁡−11z=π{\sec ^{ - 1}}\sqrt {1 + {x^2}} + \mathrm{cose}{\mathrm{c}^{ - 1}}\frac{{\sqrt {1 + {y^2}} }}{y} + {\cot ^{ - 1}}\frac{1}{z} = \pi, then x+y+zx + y + z is equal to
View answer and explanation
Correct option: A
xyzxyz
Option A (xyzxyz) is correct because the given equation can be transformed into tan⁡−1x+tan⁡−1y+tan⁡−1z=π{\tan ^{ - 1}}x + {\tan ^{ - 1}}y + {\tan ^{ - 1}}z = \pi, which implies x+y+z=xyzx + y + z = xyz.
71 Medium
71 of 221
Application +4 | -1
If sin⁡−1x−cos⁡−1x=π6,{\sin ^{ - 1}}x - {\cos ^{ - 1}}x = \frac{\pi }{6}, then xx is
View answer and explanation
Correct option: B
32\frac{{\sqrt 3 }}{2}
Option B (32\frac{{\sqrt 3 }}{2}) is correct because solving the equation using trigonometric identities leads to this value for x.
72 Medium
72 of 221
Application +4 | -1
If tan⁡−1(a/x)+tan⁡−1(b/x)=π/2\tan ^{-1}(a / x)+\tan ^{-1}(b / x)=\pi / 2, then xx is equal to
View answer and explanation
Correct option: D
ab\sqrt{a b}
Option D, ab\sqrt{ab}, is correct because by applying the tangent addition formula and using the given condition, we can derive that x=abx = \sqrt{ab}.
73 Medium
73 of 221
Application +4 | -1
If cos⁡(2sin⁡−1x)=19\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}, then xx is equal to
View answer and explanation
Correct option: C
23,−23\frac{2}{3},-\frac{2}{3}
Option C (23,−23\frac{2}{3}, -\frac{2}{3}) is correct because solving the trigonometric equation cos⁡(2sin⁡−1x)=19\cos(2\sin^{-1}x) = \frac{1}{9} yields both positive and negative solutions for x.
74 Medium
74 of 221
Analyzing +4 | -1
Sum of infinite terms of the series

cot⁡−1(12+34)+cot⁡−1(22+34)+cot⁡−1(32+34)+...{\cot ^{ - 1}}\left( {{1^2} + \frac{3}{4}} \right) + {\cot ^{ - 1}}\left( {{2^2} + \frac{3}{4}} \right) + {\cot ^{ - 1}}\left( {{3^2} + \frac{3}{4}} \right) + ... is
View answer and explanation
Correct option: B
tan⁡−1(2){\tan ^{ - 1}}\left( 2 \right)
Option B (tan⁡−1(2){\tan ^{ - 1}}( 2 )) is correct because the series telescopes to tan⁡−1(n+12)−tan⁡−1(n−12){\tan ^{ - 1}}(n + \frac{1}{2}) - {\tan ^{ - 1}}(n - \frac{1}{2}) and sums to tan⁡−1(2){\tan ^{ - 1}}(2).
75 Medium
75 of 221
Application +4 | -1
If x∈(3π2,2π)x \in\left(\frac{3 \pi}{2}, 2 \pi\right) then value of the expression sin⁡−1(cos⁡(cos⁡−1(cos⁡x)+sin⁡−1(sin⁡x)))\sin ^{-1}\left(\cos \left(\cos ^{-1}(\cos x)+\sin ^{-1}(\sin x)\right)\right) equals
View answer and explanation
Correct option: B
π2\frac{\pi}{2}
Option B (π2\frac{\pi}{2}) is correct because the expression simplifies to sin⁡−1(1)\sin^{-1}(1), which equals π2\frac{\pi}{2}.

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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
76 Medium
76 of 221
Application +4 | -1
The value of cos⁡−1x+cos⁡−1(x2+123−3x2)(1/2≤x≤1)\cos ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{1}{2} \sqrt{3-3 x^2}\right) (1 / 2 \leq x \leq 1) is equal to
View answer and explanation
Correct option: B
π/3\pi / 3
Option B (π/3\pi / 3) is correct because the expression simplifies to cos⁡−1x+(π/3−cos⁡−1x)\cos^{-1}x + (\pi/3 - \cos^{-1}x), which equals π/3\pi/3.
77 Medium
77 of 221
Application +4 | -1
If cos⁡−1[sin⁡(sin⁡−1(sin⁡x))]=0\cos ^{-1}\left[\sin \left(\sin ^{-1}(\sin x)\right)\right]=0, then the value of tan⁡x2\tan \frac{x}{2} is (x∈[0,2π])(x \in[0,2 \pi])
View answer and explanation
Correct option: A
1
Option A (1) is correct because the equation simplifies to sin x = 1, which implies x = π/2, and thus tan(x/2) = tan(π/4) = 1.
78 Medium
78 of 221
Application +4 | -1
If tan⁡−1x=2cot⁡−17+cos⁡−1(35)\tan ^{-1} x=2 \cot ^{-1} 7+\cos ^{-1}\left(\frac{3}{5}\right), then x=x=
View answer and explanation
Correct option: A
11744\frac{117}{44}
Option A (11744\frac{{117}}{{44}}) is the correct answer because it is obtained by simplifying the given trigonometric equation using the tangent addition formula.
79 Medium
79 of 221
Application +4 | -1
If cos⁡−1p+cos⁡−1q+cos⁡−1r=3π{\cos ^{ - 1}}p + {\cos ^{ - 1}}q + {\cos ^{ - 1}}r = 3\pi, then p2+q2+r2+2pqr{p^2} + {q^2} + {r^2} + 2pqr is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because the given condition cos⁡−1p+cos⁡−1q+cos⁡−1r=3π{\cos ^{ - 1}}p + {\cos ^{ - 1}}q + {\cos ^{ - 1}}r = 3\pi implies p=q=r=−1p = q = r = -1, and thus p2+q2+r2+2pqr=1{p^2} + {q^2} + {r^2} + 2pqr = 1.
80 Medium
80 of 221
Application +4 | -1
The value of sin⁡−1(1213)−sin⁡−1(35)\sin ^{-1}\left(\frac{12}{13}\right)-\sin ^{-1}\left(\frac{3}{5}\right) is equal to :
View answer and explanation
Correct option: B
π2−sin⁡−1(5665)\frac{\pi}{2}-\sin ^{-1}\left(\frac{56}{65}\right)
The correct answer is B (π2−sin⁡−1(5665)\frac{\pi}{2}-\sin^{-1}(\frac{56}{65})) because it involves using the sine subtraction formula and trigonometric identities.
81 Medium
81 of 221
Application +4 | -1
The value of cos⁡−1[cot⁡{sin⁡−1(2−34)+cos⁡−1(124)+sec⁡−1(2)}]\cos ^{-1}\left[\cot \left\{\sin ^{-1}\left(\sqrt{\frac{2-\sqrt{3}}{4}}\right)+\cos ^{-1}\left(\frac{\sqrt{12}}{4}\right)+\sec ^{-1}(\sqrt{2})\right\}\right] is equal to
View answer and explanation
Correct option: A
π2\frac{\pi}{2}
Option A (π2\frac{{\pi}}{{2}}) is correct because by simplifying the expression inside the inverse cosine function, we find that it equals zero, and the inverse cosine of zero is π2\frac{{\pi}}{{2}}.
82 Medium
82 of 221
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 using the identity for the inverse cotangent function.
83 Medium
83 of 221
Application +4 | -1
S=tan⁡−1(1n2+n+1)+tan⁡−1(1n2+3n+3)+…S=\tan ^{-1}\left(\frac{1}{n^2+n+1}\right)+\tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right)+\ldots +tan⁡−1(11+(n+19)(n+20))+\tan ^{-1}\left(\frac{1}{1+(n+19)(n+20)}\right), then tan⁡S\tan S is equal to:
View answer and explanation
Correct option: C
20n2+20n+1\frac{20}{n^2+20 n+1}
Option C (20n2+20n+1\frac{20}{n^2+20n+1}) is correct because the sum telescopes to tan⁡−1(n+20n+1)−tan⁡−1(nn+1)\tan^{-1}(\frac{n+20}{n+1}) - \tan^{-1}(\frac{n}{n+1}), and applying the tangent function simplifies to the given expression.
84 Medium
84 of 221
Application +4 | -1
The value of sin⁡−1{sin⁡[2cot⁡+−1(2−1)]}\sin ^{-1}\left\{\sin \left[2 \cot +^{-1}(\sqrt{2}-1)\right]\right\} is
View answer and explanation
Correct option: D
None of these
None of these is correct because the expression simplifies to π/4 after applying trigonometric identities and considering the range.
85 Medium
85 of 221
Application +4 | -1
The set of values of xx satisfying x2−x−π+sin⁡−1(sin⁡2)<0x^2-x-\pi+\sin ^{-1}(\sin 2)<0, is
View answer and explanation
Correct option: C
(−1,2)(-1,2)
Option C ((-1,2)) is correct because simplifying the inequality x2−x−π+sin⁡−1(sin⁡2)<0x^2 - x - \pi + \sin^{-1}(\sin 2) < 0 leads to the solution set (-1, 2).
86 Medium
86 of 221
Application +4 | -1
The value of tan⁡−1tan⁡(−6)\tan ^{-1} \tan (-6) is equal to-
View answer and explanation
Correct option: D
2π−62 \pi-6
Option D (2π−62 \pi - 6) is correct because it correctly adjusts the angle -6 to fall within the principal range of the inverse tangent function.
87 Medium
87 of 221
Application +4 | -1
Value of sin⁡−1{sin⁡[cos⁡−1(cos⁡{sin⁡−1[sin⁡(3π4)]})]}\sin ^{-1}\left\{\sin \left[\cos ^{-1}\left(\cos \left\{\sin ^{-1}\left[\sin \left(\frac{3 \pi}{4}\right)\right]\right\}\right)\right]\right\} equals
View answer and explanation
Correct option: C
π4\frac{\pi}{4}
Option C (π4\frac{\pi}{4}) is correct because simplifying the nested trigonometric and inverse trigonometric functions leads to the final value of π4\frac{\pi}{4}.
88 Medium
88 of 221
Application +4 | -1
If θ=cot⁡−17+cot⁡−18+cot⁡−118\theta=\cot ^{-1} 7+\cot ^{-1} 8+\cot ^{-1} 18, then cot⁡θ\cot \theta is equal to
View answer and explanation
Correct option: B
3
Option B (3) is correct because by applying the formula for the sum of inverse cotangents, the expression simplifies to cot⁡−1(3)\cot^{-1}(3), thus cot⁡θ=3\cot \theta = 3.
89 Medium
89 of 221
Analyzing +4 | -1
If sin⁡(2cos⁡−1(cot⁡(2tan⁡−1x)))=0,x>0\sin \left(2 \cos ^{-1}\left(\cot \left(2 \tan ^{-1} x\right)\right)\right)=0, x > 0, then which of the following is incorrect?
View answer and explanation
Correct option: C
x=3x=3
Option B (I, II, III only) is correct because the given conditions lead to (I-A)(I-B) = 0, implying that |I-A| = 0 and |I-B| = 0, and A+B = AB+I.
90 Medium
90 of 221
Application +4 | -1
If sin⁡−1x+sin⁡−1y+sin⁡−1z=          3π2  and  f(1)=2{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \frac{{\;\;\;\;\;3\pi }}{2}\;\mathrm{and}\;f\left( 1 \right) = 2,

f(p+q)=f(p).f(q),∀  p,  q∈R,  f\left( {p + q} \right) = f\left( p \right).f\left( q \right),\forall \;p,\;q \in R,\; then

xf(1)+yf(2)+zf(3)−(x+y+z)xf(1)+yf(2)+zf(3)  {x^{f\left( 1 \right)}} + {y^{f\left( 2 \right)}} + {z^{f\left( 3 \right)}} - \frac{{\left( {x + y + z} \right)}}{{{x^{f\left( 1 \right)}} + {y^{f\left( 2 \right)}} + {z^{f\left( 3 \right)}}}}\; is equal to
View answer and explanation
Correct option: C
2
Option C (2) is correct because given the conditions, x=y=z=1, f(1)=2, f(2)=4, f(3)=8, which simplifies the expression to 3 - 1 = 2.
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