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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Mixed difficulty

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511 Easy
511 of 605
Analyzing +4 | -1
The sum of the infinite series

sin⁡−1(12)+sin⁡−1(2−16)+sin⁡−1(3−212)+...{\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) + {\sin ^{ - 1}}\left( {\frac{{\sqrt 2 - 1}}{{\sqrt 6 }}} \right) + {\sin ^{ - 1}}\left( {\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt {12} }}} \right) + ...

+...+sin⁡−1(n−(n−1){n(n+1)})+...+ ... + {\sin ^{ - 1}}\left( {\frac{{\sqrt n - \sqrt {\left( {n - 1} \right)} }}{{\sqrt {\left\{ {n\left( {n + 1} \right)} \right\}} }}} \right) + ... is
View answer and explanation
Correct option: C
π2\frac{\pi }{2}
Option C (π2\frac{\pi}{2}) is correct because the series telescopes to tan⁡−1(n){\tan^{-1}}(\sqrt{n}) which approaches π2\frac{\pi}{2} as n approaches infinity.
512 Easy
512 of 605
Application +4 | -1
The value of sin (cot⁡−1x)({\cot ^{ - 1}}x) is
View answer and explanation
Correct option: D
(1+x2)−1/2{\left( {1 + {x^2}} \right)^{ - 1/2}}
Option D ((1+x2)−1/2{\left( {1 + {x^2}} \right)^{ - 1/2}}) is correct because by using trigonometric identities and the definition of cotangent, we can express sin⁡(cot⁡−1x)\sin({\cot ^{ - 1}}x) as (1+x2)−1/2{\left( {1 + {x^2}} \right)^{ - 1/2}}.
513 Easy
513 of 605
Application +4 | -1
If tan⁡−1x+tan⁡−1y+tan⁡−1z=π\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\pi, then the value of x+y+z−xyzx+y+z-x y z is
View answer and explanation
Correct option: A
0
Option A (0) is correct because if tan⁡−1x+tan⁡−1y+tan⁡−1z=π\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi, then x+y+z=xyzx + y + z = xyz, which implies x+y+z−xyz=0x + y + z - xyz = 0.
514 Easy
514 of 605
Application +4 | -1
If (tan⁡−1x)2+(cot⁡−1x)2=5π28{\left( {{{\tan }^{ - 1}}x} \right)^2} + {\left( {{{\cot }^{ - 1}}x} \right)^2} = \frac{{5{\pi ^2}}}{8}, then xx is equal to
View answer and explanation
Correct option: D
−1- 1
Option D (-1) is correct because solving the equation using the property (tan-1x)2 + (cot-1x)2 = 5pi2/8 leads to x = -1.
515 Easy
515 of 605
Analyzing +4 | -1
−2π5-\frac{2 \pi}{5} is the principal value of
View answer and explanation
Correct option: B
sin⁡−1(sin⁡7π5)\sin ^{-1}\left(\sin \frac{7 \pi}{5}\right)
Option B is correct because sin⁡−1(sin⁡x)\sin^{-1}(\sin x) has a principal value range of [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}], and sin⁡−1(sin⁡7π5)=−2π5\sin^{-1}(\sin \frac{7\pi}{5}) = -\frac{2\pi}{5}.
516 Medium
516 of 605
Application +4 | -1
cos⁡−1(1517)+2tan⁡−1(15)\cos ^{-1}\left(\frac{15}{17}\right)+2 \tan ^{-1}\left(\frac{1}{5}\right) is equal to
View answer and explanation
Correct option: B
sin⁡−1(171221)\sin ^{-1}\left(\frac{171}{221}\right)
Option B (sin⁡−1(171221)\sin^{-1}(\frac{{171}}{{221}})) is correct because simplifying the expression using trigonometric identities results in this value.
517 Medium
517 of 605
Analyzing +4 | -1
cot⁡−1(cos⁡α)−tan⁡−1(cos⁡α)=x\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x, then sin⁡x=\sin x=
View answer and explanation
Correct option: A
tan⁡2(α2)\tan ^{2}\left(\frac{\alpha}{2}\right)
Option A (tan⁡2(α2)\tan ^{2}\left(\frac{{\alpha}}{{2}}\right)) is correct because by using trigonometric identities and simplifying the expression, we find that sin⁡x=tan⁡2(α2)\sin x = \tan^2(\frac{\alpha}{2}).
518 Easy
518 of 605
Application +4 | -1
If tan⁡−1(x+2)+tan⁡−1(x−2)−tan⁡−1(12)=0,  {\tan ^{ - 1}}\left( {x + 2} \right) + {\tan ^{ - 1}}\left( {x - 2} \right) - {\tan ^{ - 1}}\left( {\frac{1}{2}} \right) = 0,\; then one of the values of xx is equal to
View answer and explanation
Correct option: D
1
Option D (1) is correct because solving the equation using the tangent addition formula gives x = 1 or x = -5, and x = 1 satisfies the original equation.
519 Medium
519 of 605
Application +4 | -1
If cos⁡−1x−cos⁡−1y2=α\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha, where −1≤x≤1,−2≤y≤2-1 \leq x \leq 1,-2 \leq y \leq 2, x≤y2x \leq \frac{y}{2}, then for all x,y,4x2−4xycos⁡α+y2x, y, 4 x^{2}-4 x y \cos \alpha+y^{2} is equal to:
View answer and explanation
Correct option: A
4sin⁡2α4 \sin ^{2} \alpha
The correct answer is A (4sin⁡2α4 \sin^2 \alpha) because it involves using the cosine subtraction formula and trigonometric identities to simplify the expression.
520 Medium
520 of 605
Application +4 | -1
If cos⁡−1p+cos⁡−1q+cos⁡−1r=π\cos ^{-1} \mathrm{p}+\cos ^{-1} \mathrm{q}+\cos ^{-1} \mathrm{r}=\pi. Then value of p2+q2\mathrm{p}^2+\mathrm{q}^2 +r2+2pqr+4+r^2+2 p q r+4 is equal to-
View answer and explanation
Correct option: C
5
Option C (5) is correct because the given condition leads to the identity p2+q2+r2+2pqr=1p^2 + q^2 + r^2 + 2pqr = 1, and adding 4 to both sides results in 5.
521 Easy
521 of 605
Analyzing +4 | -1
If a>b>0a > b > 0, then the value of tan⁡−1(ab)+tan⁡−1(a+ba−b){\tan ^{ - 1}}\left( {\frac{a}{b}} \right) + {\tan ^{ - 1}}\left( {\frac{{a + b}}{{a - b}}} \right) depends on
View answer and explanation
Correct option: D
Neither aa nor bb
Option D (Neither aa nor bb) is correct because the expression simplifies to a constant value independent of aa and bb.
522 Easy
522 of 605
Application +4 | -1
If sin⁡−1(2x1−x2)−2sin⁡−1x=0{\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right) - 2{\sin ^{ - 1}}x = 0, then xx belongs to the interval
View answer and explanation
Correct option: B
[−1/2,  1/2]\left[ { - 1/\sqrt 2 ,\;1/\sqrt 2 } \right]
Option D ([1/2,  1]\left[ {1/\sqrt 2 ,\;1} \right]) is incorrect. Option B ([−1/2,  1/2]\left[ { - 1/\sqrt 2 ,\;1/\sqrt 2 } \right]) is correct because the equation holds true when xx is within this interval, based on the properties of inverse trigonometric functions.
523 Easy
523 of 605
Application +4 | -1
If tan⁡−1[3sin⁡2α5+3cos⁡2α]+tan⁡−1[tan⁡α4]=λα4\tan ^{-1}\left[\frac{3 \sin 2 \alpha}{5+3 \cos 2 \alpha}\right]+\tan ^{-1}\left[\frac{\tan \alpha}{4}\right]=\frac{\lambda \alpha}{4}; where −π2<α<π2-\frac{\pi}{2}<\alpha<\frac{\pi}{2}, then find λ\lambda
View answer and explanation
Correct option: A
4
Option A (4) is correct because simplifying the given expression leads to α = λα / 4, which implies λ = 4.
524 Medium
524 of 605
Analyzing +4 | -1
sin⁡−1(sin⁡3x2−12x2+2)<34\sin ^{-1}\left(\sin \frac{3 x^2-1}{2 x^2+2}\right) < \frac{3}{4} if
View answer and explanation
Correct option: A
−53<x<53-\sqrt{\frac{5}{3}} < x < \sqrt{\frac{5}{3}}
Option C (1/e) is correct because by applying L'Hopital's rule to the logarithm of the expression, we find the limit to be -1, thus the original limit is 1/e.
525 Medium
525 of 605
Application +4 | -1
sec⁡2(tan⁡−12)+cosec⁡2(cot⁡−13)\sec ^2\left(\tan ^{-1} 2\right)+\operatorname{cosec}^2\left(\cot ^{-1} 3\right) is equal to
View answer and explanation
Correct option: D
15
Option D (15) is correct because sec⁡2(tan⁡−12)+cosec⁡2(cot⁡−13)=(1+tan⁡2(tan⁡−12))+(1+cot⁡2(cot⁡−13))=1+4+1+9=15\sec^2(\tan^{-1} 2) + \operatorname{cosec}^2(\cot^{-1} 3) = (1 + \tan^2(\tan^{-1} 2)) + (1 + \cot^2(\cot^{-1} 3)) = 1 + 4 + 1 + 9 = 15.

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526 Easy
526 of 605
Application +4 | -1
If AA=tan⁡−1(x32k−x)  {\tan ^{ - 1}}\left( {\frac{{x\sqrt 3 }}{{2k - x}}} \right)\; and B=tan⁡−1(2x−kk3),B = {\tan ^{ - 1}}\left( {\frac{{2x - k}}{{k\sqrt 3 }}} \right), then the value of A−BA - B is
View answer and explanation
Correct option: D
30∘30^\circ
Option D (30∘30^\circ) is correct because by using the formula for tan⁡(A−B)\tan(A-B) and simplifying, we find that A−B=30∘A - B = 30^\circ.
527 Medium
527 of 605
Application +4 | -1
The value of cos⁡(2cos⁡−1x+sin⁡−1x)  at  x=15  \cos (2{\cos ^{ - 1}}x + {\sin ^{ - 1}}x)\;\mathrm{at}\;x = \frac{1}{5}\; is
View answer and explanation
Correct option: D
−265- \frac{{2\sqrt 6 }}{5}
Option D (−265- \frac{{2\sqrt 6 }}{5}) is correct because the expression simplifies using trigonometric identities to this value when x = 1/5.
528 Easy
528 of 605
Application +4 | -1
The value of sin⁡(cot⁡−1(cos⁡(tan⁡−1x)))\sin \left(\cot ^{-1}\left(\cos \left(\tan ^{-1} x\right)\right)\right) is
View answer and explanation
Correct option: B
x2+1x2+2\sqrt{\frac{x^2+1}{x^2+2}}
Option B, x2+1x2+2\sqrt{\frac{x^2+1}{x^2+2}}, is correct because by using trigonometric substitutions and identities, the expression can be simplified to this form.
529 Medium
529 of 605
Application +4 | -1
cos⁡[2tan⁡−1(17)]\cos \left[2 \tan ^{-1}\left(\frac{1}{7}\right)\right] equals
View answer and explanation
Correct option: A
sin⁡(4cot⁡−13)\sin \left(4 \cot ^{-1} 3\right)
Option A (sin⁡(4cot⁡−13)\sin \left(4 \cot ^{-1} 3\right)) is correct because by applying trigonometric identities, both expressions simplify to 2425\frac{{24}}{{25}}.
530 Easy
530 of 605
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 it simplifies to tan⁡(2tan⁡−1a)\tan(2 \tan^{-1} a) using trigonometric identities, which further simplifies to the given expression.
531 Medium
531 of 605
Application +4 | -1
Let tan⁡−1y=tan⁡−1x+tan⁡−1(2x1−x2)\tan ^{-1} y=\tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right), where or ∣x∣<13|x|<\frac{1}{\sqrt{3}}. Then a value of yy is :
View answer and explanation
Correct option: A
3x−x31−3x2\frac{3 x-x^{3}}{1-3 x^{2}}
The correct answer is A (3x−x31−3x2\frac{3 x-x^{3}}{1-3 x^{2}}) because it involves using the triple angle formula for tangent.
532 Medium
532 of 605
Application +4 | -1
If x∈(1,  ∞)x \in \left( {1,\;\infty } \right), then sin⁡−1(2x1+x2){\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right) equals
View answer and explanation
Correct option: B
π−2tan⁡−1x\pi - 2{\tan ^{ - 1}}x
Option B (π−2tan⁡−1x\pi - 2{\tan ^{ - 1}}x) is correct because when x∈(1,∞)x \in (1, \infty), the range of 2tan⁡−1x2\tan^{-1}x is (π2,π)(\frac{\pi}{2}, \pi), and we need to adjust the result of sin⁡−1(sin⁡(2θ))\sin^{-1}(\sin(2\theta)) to fall within the principal range of sin⁡−1\sin^{-1}.
533 Easy
533 of 605
Application +4 | -1
If 3sin⁡−12x1+x2−4cos⁡−11+x1+x2+2tan⁡−12x1−x2=π3,3{\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} - 4{\cos ^{ - 1}}\frac{{1 + x}}{{1 + {x^2}}} + 2{\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} = \frac{\pi }{3}, then value of xx is
View answer and explanation
Correct option: B
13\frac{1}{{\sqrt 3 }}
Option B (13\frac{1}{\sqrt{3}}) is correct because it is the solution obtained by substituting x=tan⁡θx = \tan \theta and simplifying the given equation using trigonometric identities.
534 Easy
534 of 605
Application +4 | -1
The value of cos⁡−1(−12){\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) among the following, is
View answer and explanation
Correct option: B
8π3\frac{{8\pi }}{3}
Option B (8π3\frac{{8\pi }}{3}) is correct because it is one of the possible values of cos⁡−1(−12){\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) obtained by considering the general solution of the cosine function.
535 Easy
535 of 605
Application +4 | -1
If sin⁡−1(6x1+9x2)=2tan⁡−1(ax)\sin ^{-1}\left(\frac{6 x}{1+9 x^2}\right)=2 \tan ^{-1}(a x), then a=a=
View answer and explanation
Correct option: A
3
Option A (3) is correct because by using the double angle formula for sine and comparing the expressions, a is found to be 3.
536 Easy
536 of 605
Application +4 | -1
cos⁡−1(1517)+2tan⁡−1(15)={\cos ^{ - 1}}\left( {\frac{{15}}{{17}}} \right) + 2{\tan ^{ - 1}}\left( {\frac{1}{5}} \right) =
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the expression simplifies to cos⁡−1(140221){\cos ^{ - 1}}(\frac{{140}}{{221}}), which is not equal to any of the other options.
537 Easy
537 of 605
Application +4 | -1
If cos⁡−135−sin⁡−145=cos⁡−1x,{\cos ^{ - 1}}\frac{3}{5} - {\sin ^{ - 1}}\frac{4}{5} = {\cos ^{ - 1}}x, then xx is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because cos⁡−135−sin⁡−145{\cos ^{ - 1}}\frac{3}{5} - {\sin ^{ - 1}}\frac{4}{5} simplifies to 0, and cos⁡−1x=0{\cos ^{ - 1}}x = 0 implies x=1x = 1.
538 Easy
538 of 605
Application +4 | -1
If −∞<x≤0- \infty < x \le 0, then cos⁡−1(1−x21+x2){\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right) equals
View answer and explanation
Correct option: B
−2tan⁡−1x- 2{\tan ^{ - 1}}x
Option B (−2tan⁡−1x- 2{\tan ^{ - 1}}x) is correct because it simplifies the given expression using trigonometric identities and the given condition on x.
539 Easy
539 of 605
Analyzing +4 | -1
The value of ′a′'a' for which ax2+sin⁡−1(x2−2x+2)+cos⁡−1(x2−2x+2)=0a{x^2} + {\sin ^{ - 1}}\left( {{x^2} - 2x + 2} \right) + {\cos ^{ - 1}}\left( {{x^2} - 2x + 2} \right) = 0 has a real solution, is
View answer and explanation
Correct option: C
−π2- \frac{\pi }{2}
Option C (−π2- \frac{\pi }{2}) is correct because substituting x=1 into the equation and solving for 'a' yields a = -π/2.
540 Easy
540 of 605
Application +4 | -1
cos⁡−112+2sin⁡−112\cos ^{-1} \frac{1}{2}+2 \sin ^{-1} \frac{1}{2} is equal to
View answer and explanation
Correct option: D
2π/32 \pi / 3
Option D (2π/32 \pi / 3) is correct because it is the result of evaluating the given expression using the principal values of the inverse trigonometric functions.
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