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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Easy

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241 Easy
241 of 365
Application +4 | -1
If sin⁡−1α+sin⁡−1β+sin⁡−1γ=3π2,  {\sin ^{ - 1}}\mathrm{\alpha } + {\sin ^{ - 1}}\mathrm{\beta } + {\sin ^{ - 1}}\mathrm{\gamma } = \frac{{3\pi }}{2},\; then αβ+αγ+βγ\mathrm{\alpha \beta } + \mathrm{\alpha \gamma } + \mathrm{\beta \gamma } is equal to
View answer and explanation
Correct option: C
3
Option C (3) is correct because, given sin⁡−1α+sin⁡−1β+sin⁡−1γ=3π2{\sin ^{ - 1}}\mathrm{\alpha } + {\sin ^{ - 1}}\mathrm{\beta } + {\sin ^{ - 1}}\mathrm{\gamma } = \frac{{3\pi }}{2}, it implies α=β=γ=1\alpha = \beta = \gamma = 1, and thus αβ+αγ+βγ=3\mathrm{\alpha \beta } + \mathrm{\alpha \gamma } + \mathrm{\beta \gamma } = 3.
242 Easy
242 of 365
Application +4 | -1
2cot⁡−1(7)+cos⁡−1(3/5)2 \cot ^{-1}(7)+\cos ^{-1}(3 / 5), in principal value is equal to
View answer and explanation
Correct option: B
cot⁡−1(44117)\cot ^{-1}\left(\frac{44}{117}\right)
Option B (cot⁡−1(44117)\cot^{-1}(\frac{44}{117})) is correct because by using the double angle formula for the inverse tangent and then the sum formula for inverse tangents, we can simplify the expression to this form.
243 Easy
243 of 365
Remembering +4 | -1
Which of the following is the principal value branch of cos⁡−1x?\cos ^{-1} \mathrm{x} ?
View answer and explanation
Correct option: C
[0,π][0, \pi]
Option C ([0,π][0, \pi]) is correct because the principal value branch of cos⁡−1x\cos^{-1} x is [0,π][0, \pi].
244 Easy
244 of 365
Application +4 | -1
The value of tan⁡−1(1)+tan⁡−1(0)+tan⁡−1(2)+tan⁡−1(3)  {\tan ^{ - 1}}\left( 1 \right) + {\tan ^{ - 1}}\left( 0 \right) + {\tan ^{ - 1}}\left( 2 \right) + {\tan ^{ - 1}}\left( 3 \right)\; is equal to
View answer and explanation
Correct option: A
π\pi
Option A (π\pi) is correct because by applying the tan⁡−1x+tan⁡−1y\tan^{-1}x + \tan^{-1}y formula and simplifying, the expression evaluates to π\pi.
245 Easy
245 of 365
Application +4 | -1
If cos⁡−1x+cos⁡−1y+cos⁡−1z=3π,  {\cos ^{ - 1}}x + {\cos ^{ - 1}}y + {\cos ^{ - 1}}z = 3\pi ,\; then xy+yz+zxxy + yz + zx is equal to
View answer and explanation
Correct option: C
3
Option C (3) is correct because when the sum of inverse cosines equals 3π3\pi, each inverse cosine must equal π\pi, leading to x = y = z = -1, and thus xy + yz + zx = 3.
246 Easy
246 of 365
Analyzing +4 | -1
If cot⁡−1(α)=cot⁡−12+cot−18+cot−118+cot⁡−132+.....{\cot ^{ - 1}}(\alpha ) = {\cot ^{ - 1}}\mathrm{2 } + \mathrm{ co}{\mathrm{t}^{ - 1}}\mathrm{8 } + \mathrm{ co}{\mathrm{t}^{ - 1}}18 + {\cot ^{ - 1}}32 + ..... upto 100 terms, then α\alpha is
View answer and explanation
Correct option: A
1.01
Option A (1.01) is correct because the sum of the series can be simplified using the arctangent subtraction formula, leading to α=1.01\alpha = 1.01.
247 Easy
247 of 365
Application +4 | -1
If tan⁡−12x+tan⁡−13x=π4\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}, then x=x=
View answer and explanation
Correct option: C
1/61 / 6
Option C (1/61 / 6) is correct because solving the equation tan⁡−12x+tan⁡−13x=π4\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4} leads to the solution x=1/6x = 1/6.
248 Easy
248 of 365
Application +4 | -1
sin⁡−11213+cos⁡−145+tan⁡−16316=\sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}=
View answer and explanation
Correct option: C
π\pi
Option C (π\pi) is correct because the sum of the inverse trigonometric functions simplifies to π\pi after applying trigonometric identities.
249 Easy
249 of 365
Application +4 | -1
If sin⁡−1x=tan⁡−1y\sin ^{-1} x=\tan ^{-1} y, what is the value of 1x2−1y2\frac{1}{x^2}-\frac{1}{y^2} ?
View answer and explanation
Correct option: A
1
Option A (1) is correct because by substituting sin⁡−1x=tan⁡−1y=θ\sin^{-1}x = \tan^{-1}y = \theta and using trigonometric identities, the expression simplifies to 1.
250 Easy
250 of 365
Analyzing +4 | -1
The solution set of the equation tan⁡−1x−cot⁡−1x=cos⁡−1(2−x)is{\tan ^{ - 1}}x - {\cot ^{ - 1}}x = {\cos ^{ - 1}}\left( {2 - x} \right)\mathrm{is}
View answer and explanation
Correct option: C
[1,3]
Option C [1,3] is the correct solution set because it satisfies the domain restrictions of the inverse trigonometric functions and the given equation.
251 Easy
251 of 365
Application +4 | -1
Principal value of cosec⁡−1(−23)\operatorname{cosec}^{-1}\left(\frac{-2}{\sqrt{3}}\right) is equal to
View answer and explanation
Correct option: A
−π3-\frac{\pi}{3}
Option A (−π3-\frac{{\pi}}{{3}}) is correct because it is the principal value of the inverse cosecant function for the given value.
252 Easy
252 of 365
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}.
253 Easy
253 of 365
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.
254 Easy
254 of 365
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 }}.
255 Easy
255 of 365
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.

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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
256 Easy
256 of 365
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.
257 Easy
257 of 365
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.
258 Easy
258 of 365
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.
259 Easy
259 of 365
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}.
260 Easy
260 of 365
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.
261 Easy
261 of 365
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}.
262 Easy
262 of 365
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.
263 Easy
263 of 365
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.
264 Easy
264 of 365
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.
265 Easy
265 of 365
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}.
266 Easy
266 of 365
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}.
267 Easy
267 of 365
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}.
268 Easy
268 of 365
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.
269 Easy
269 of 365
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}.
270 Easy
270 of 365
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}.
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