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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Mixed difficulty

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Showing 421-450 of 605 questions

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421 Medium
421 of 605
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.
422 Easy
422 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
2π3\frac{{2\pi }}{3}
Option D (2π3\frac{{2\pi }}{3}) is correct because substituting the known values of inverse cosine and inverse sine simplifies to 2π3\frac{{2\pi }}{3}.
423 Hard
423 of 605
Understanding +4 | -1
cos⁡−1a−xa−b=sin⁡−1x−ba−b\cos ^{-1} \sqrt{\frac{a-x}{a-b}}=\sin ^{-1} \sqrt{\frac{x-b}{a-b}} is possible if
View answer and explanation
Correct option: A
a>x>ba > x > b
Option A (a>x>ba > x > b) is correct because it satisfies the conditions for the inverse trigonometric functions to be defined and equal.
424 Easy
424 of 605
Analyzing +4 | -1
∑m=1ntan⁡−1(2mm4+m2+2)\sum \limits_{m = 1}^n {\tan ^{ - 1}}\left( {\frac{{2m}}{{{m^4} + {m^2} + 2}}} \right) is equal to
View answer and explanation
Correct option: A
tan⁡−1(n2+nn4+n2+2){\tan ^{ - 1}}\left( {\frac{{{n^2} + n}}{{{n^4} + {n^2} + 2}}} \right)
Option D is incorrect. Option A (tan⁡−1(n2+nn4+n2+2){\tan ^{ - 1}}\left( {\frac{{{{{n^2} + n}}}}{{{{{n^4} + {n^2} + 2}}}}} \right)) is the correct answer because the sum telescopes, leaving a simplified expression involving nn.
425 Easy
425 of 605
Application +4 | -1
If tan⁡−11−x1+x=12tan⁡−1x{\tan ^{ - 1}}\frac{{1 - x}}{{1 + x}} = \frac{1}{2}{\tan ^{ - 1}}x, then the 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 solving the equation tan⁡−11−x1+x=12tan⁡−1x{\tan ^{ - 1}}\frac{{1 - x}}{{1 + x}} = \frac{1}{2}{\tan ^{ - 1}}x leads to x=13x = \frac{1}{{\sqrt 3 }}.
426 Easy
426 of 605
Analyzing +4 | -1
If [cot⁡−1x]+[cos⁡−1x]=0,  \left[ {{{\cot }^{ - 1}}x} \right] + \left[ {{{\cos }^{ - 1}}x} \right] = 0,\; where xx is a non-negative real number and [.] denotes the greatest integer function, then complete set of values of xx is
View answer and explanation
Correct option: B
(cot⁡1,1)\left( {\cot 1,1} \right)
The correct answer is B, (cot⁡1,1)\left( {\cot 1,1} \right), because the given condition implies that both inverse cotangent and inverse cosine must have a greatest integer value of 0, leading to the specified interval.
427 Hard
427 of 605
Application +4 | -1
If sin⁡−1(−x22+x34−…)+cos⁡−1(x2−x42+x64−…)\sin ^{-1}\left(-\frac{x^2}{2}+\frac{x^3}{4}-\ldots\right)+\cos ^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\ldots\right) =π2=\frac{\pi}{2} for 0<∣x∣<20<|x|<\sqrt{2}, then xx equals
View answer and explanation
Correct option: B
1
Option B (1) is correct because simplifying the given equation using the properties of inverse trigonometric functions and geometric series leads to x = 1.
428 Easy
428 of 605
Application +4 | -1
If sin⁡−1x+sin⁡−1(1−x)=cos⁡−1x{\sin ^{ - 1}}x + {\sin ^{ - 1}}\left( {1 - x} \right) = {\cos ^{ - 1}}x, then xx equals
View answer and explanation
Correct option: C
0,120,\frac{1}{2}
Option C (0,120,\frac{1}{2}) is the correct answer because these values satisfy the given inverse trigonometric equation.
429 Medium
429 of 605
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.
430 Easy
430 of 605
Application +4 | -1
If sin⁡−1x+sin⁡−1y=2π/3\sin ^{-1} x+\sin ^{-1} y=2 \pi / 3, then cos⁡−1x+cos⁡−1y\cos ^{-1} x+\cos ^{-1} y is equal to
View answer and explanation
Correct option: B
π/3\pi / 3
Option B (π/3\pi / 3) is correct because using the identity sin⁡−1x+cos⁡−1x=π/2\sin^{-1}x + \cos^{-1}x = \pi/2, we can rewrite the expression and solve for cos⁡−1x+cos⁡−1y\cos^{-1}x + \cos^{-1}y.
431 Medium
431 of 605
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.
432 Easy
432 of 605
Application +4 | -1
cos⁡{cos⁡−1(−17)+sin⁡−1(−17)}  =\cos \left\{ {{{\cos }^{ - 1}}\left( { - \frac{1}{7}} \right) + {{\sin }^{ - 1}}\left( { - \frac{1}{7}} \right)} \right\}\; =
View answer and explanation
Correct option: B
00
Option B (0) is correct because cos⁡−1(−17)+sin⁡−1(−17)=π2{\cos ^{ - 1}}( - \frac{1}{7}) + {\sin ^{ - 1}}( - \frac{1}{7}) = \frac{\pi}{2}, and cos⁡(π2)=0\cos(\frac{\pi}{2}) = 0.
433 Easy
433 of 605
Analyzing +4 | -1
The value of ∑r=0∞tan⁡−1(11+r+r2)  \sum \limits_{r = 0}^\infty {\tan ^{ - 1}}\left( {\frac{1}{{1 + r + {r^2}}}} \right)\; is equal to
View answer and explanation
Correct option: A
π2\frac{\pi }{2}
Option A (π2\frac{\pi }{2}) is correct because the summation telescopes, leaving only tan⁡−1(∞){\tan ^{ - 1}}(\infty ), which equals π2\frac{\pi }{2}.
434 Easy
434 of 605
Application +4 | -1
sin⁡(2sin⁡−16365)\sin \left( {2{{\sin }^{ - 1}}\sqrt {\frac{{63}}{{65}}} } \right) is equal to
View answer and explanation
Correct option: A
212665\frac{{2\sqrt {126} }}{{65}}
Option A is correct because it is the result of applying the double angle formula for sine and simplifying the expression.
435 Easy
435 of 605
Application +4 | -1
If θ=tan⁡−1a,ϕ\mathrm{\theta } = {\tan ^{ - 1}}a,\phi =tan⁡−1b  and  ab{\tan ^{ - 1}}b\;\mathrm{and}\;ab=−1,- 1,then (θ−ϕ)\mathrm{\theta } - \phi ) is equal to
View answer and explanation
Correct option: C
π2\frac{\pi }{2}
Option C (π2\frac{\pi }{2}) is correct because given ab=−1ab = -1, we can deduce that tan⁡θ=−cot⁡ϕ\tan{\theta} = -\cot{\phi}, which leads to θ−ϕ=π2\theta - \phi = \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
436 Medium
436 of 605
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.
437 Easy
437 of 605
Application +4 | -1
The value of tan⁡{12cos⁡−1(53)}\tan \left\{ {\frac{1}{2}{{\cos }^{ - 1}}\left( {\frac{{\sqrt 5 }}{3}} \right)} \right\}, is
View answer and explanation
Correct option: C
12(3−5)\frac{1}{2}\left( {3 - \sqrt 5 } \right)
Option C (12(3−5)\frac{1}{2}\left( {3 - \sqrt 5 } \right)) is correct because by using the half-angle formula for tangent, we can simplify the expression and find the value.
438 Easy
438 of 605
Application +4 | -1
If α=13sin⁡−1(2x1+x2)+13cos⁡−1(1−x21+x2)\alpha=\frac{1}{3} \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\frac{1}{3} \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) where x≥43x \geq \frac{4}{3}, then the value of cos⁡2α+sec⁡α+333\frac{\cos 2 \alpha+\sec \alpha+3 \sqrt{3}}{\sqrt{3}} is equal to:
View answer and explanation
Correct option: D
(32+3)\left(\frac{\sqrt{3}}{2}+3\right)
Option D ((32+3)\left(\frac{\sqrt{{3}}}{{2}}+3\right)) is correct because after simplifying the expression and substituting α=π3\alpha = \frac{\pi}{3}, the value of the expression is 32+3\frac{\sqrt{{3}}}{{2}}+3.
439 Easy
439 of 605
Application +4 | -1
If tan⁡−1x−tan⁡−1y=tan⁡−1A,then  A  is  equal  to{\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}A,\mathrm{then\;}A\mathrm{\;is\;equal\;to}
View answer and explanation
Correct option: C
x−y1+xy\frac{{x - y}}{{1 + xy}}
Option C (x−y1+xy\frac{{x - y}}{{1 + xy}}) is correct because it directly applies the tangent subtraction formula for inverse trigonometric functions.
440 Medium
440 of 605
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).
441 Easy
441 of 605
Analyzing +4 | -1
The sum of series

tan⁡−111+1+12+tan⁡−111+2+22+tan⁡−111+3+32+…{\tan ^{ - 1}}\frac{1}{{1 + 1 + {1^2}}} + {\tan ^{ - 1}}\frac{1}{{1 + 2 + {2^2}}} + {\tan ^{ - 1}}\frac{1}{{1 + 3 + {3^2}}} + \ldots

∞\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 given series telescopes to tan⁡−1(∞)−tan⁡−1(1){\tan ^{ - 1}}(\infty ) - {\tan ^{ - 1}}(1), which simplifies to π2−π4=π4\frac{\pi }{2} - \frac{\pi }{4} = \frac{\pi }{4}.
442 Easy
442 of 605
Application +4 | -1
tan⁡[π2+12cos⁡−1(ab)]+tan⁡[π4−12cos⁡−1(ab)]\tan \left[ {\frac{\pi }{2} + \frac{1}{2}{{\cos }^{ - 1}}\left( {\frac{a}{b}} \right)} \right] + \tan \left[ {\frac{\pi }{4} - \frac{1}{2}{{\cos }^{ - 1}}\left( {\frac{a}{b}} \right)} \right] is equal to
View answer and explanation
Correct option: B
2ba\frac{{2b}}{a}
Option B (2ba\frac{{2b}}{a}) is correct because by using trigonometric identities and simplifying the expression, it reduces to 2cos⁡2ϕ\frac{2}{{\cos 2\phi}}, which equals 2ba\frac{{2b}}{a}.
443 Easy
443 of 605
Application +4 | -1
sin⁡−145+2tan⁡−113{\sin ^{ - 1}}\frac{4}{5} + 2{\tan ^{ - 1}}\frac{1}{3} is equal to
View answer and explanation
Correct option: C
π2\frac{\pi }{2}
Option C (π2\frac{\pi}{2}) is the correct answer because the expression simplifies to sin⁡−145+cos⁡−145{\sin ^{ - 1}}\frac{4}{5} + {\cos ^{ - 1}}\frac{4}{5}, which equals π2\frac{\pi}{2}.
444 Easy
444 of 605
Application +4 | -1
If 2sin⁡−1x=sin⁡−1(2x1−x2),{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right), then xx is equal to
View answer and explanation
Correct option: C
[−12,12]\left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]
Option C ([−12,12]\left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]) is correct because the range of xx is determined by the domain restrictions of the inverse sine function.
445 Medium
445 of 605
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.
446 Medium
446 of 605
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).
447 Easy
447 of 605
Understanding +4 | -1
cos⁡−1(12)+2sin⁡−1(12){\cos ^{ - 1}}\left( {\frac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\frac{1}{2}} \right) is equal to
View answer and explanation
Correct option: C
2π3\frac{{2\pi }}{3}
Option C (2π3\frac{{2\pi }}{3}) is correct because cos⁡−1(12)=π3{\cos ^{ - 1}}(\frac{1}{2}) = \frac{\pi}{3} and sin⁡−1(12)=π6{\sin ^{ - 1}}(\frac{1}{2}) = \frac{\pi}{6}, so the expression evaluates to π3+2(π6)=2π3\frac{\pi}{3} + 2(\frac{\pi}{6}) = \frac{{2\pi }}{3}.
448 Easy
448 of 605
Analyzing +4 | -1
If A   and   BA\,\;\mathrm{and}\;\,B are acute positive angles satisfying the equations 3  sin⁡2A+2  sin⁡2  B=1   and   3  sin⁡  2A−2  sin⁡  2B=03\,\,{\sin ^2}A + 2\,\,{\sin ^2}\,\,B = 1\,\;\mathrm{and}\;\,3\,\,\sin \,\,2A - 2\,\,\sin \,\,2B = 0, then A+2BA + 2B is equal to
View answer and explanation
Correct option: B
π2\frac{\pi }{2}
Option B (π2\frac{\pi}{2}) is correct because, by manipulating the given trigonometric equations and using trigonometric identities, we can find that A+2B=π2A + 2B = \frac{\pi}{2}.
449 Easy
449 of 605
Application +4 | -1
The value of cot⁡−19+cosec−1414{\cot ^{ - 1}}9 + \mathrm{cose}{\mathrm{c}^{ - 1}}\frac{{\sqrt {41} }}{4} is
View answer and explanation
Correct option: B
π4\frac{\pi }{4}
Option B (π4\frac{\pi}{4}) is correct because by converting to arctangent and using the arctangent addition formula, the expression simplifies to tan⁡−1(1)\tan^{-1}(1).
450 Easy
450 of 605
Application +4 | -1
tan⁡−1(1/4)+tan⁡−1(2/9)\tan ^{-1}(1 / 4)+\tan ^{-1}(2 / 9) is equal to
View answer and explanation
Correct option: A
(1/2)cos⁡−1(3/5)(1 / 2) \cos ^{-1}(3 / 5)
Option A ((1/2)cos⁡−1(3/5)(1 / 2) \cos ^{-1}(3 / 5)) is correct because the sum of the two inverse tangent functions simplifies to tan⁡−1(1/2)\tan^{-1}(1/2), which is equal to (1/2)cos⁡−1(3/5)(1/2)\cos^{-1}(3/5).
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