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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Mixed difficulty

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Showing 391-420 of 605 questions

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391 Medium
391 of 605
Application +4 | -1
If x>0,y>0x > 0, y > 0 and x>yx > y, then tan⁡−1(x/y)+\tan ^{-1}(x / y)+ tan⁡−1[(x+y)/(x−y)]\tan ^{-1}[(x+y) /(x-y)] is equal to
View answer and explanation
Correct option: C
3π/43 \pi / 4
Option C (3π/4) is correct because using the formula for the sum of two arctangents and considering the given conditions, we arrive at the correct answer.
392 Easy
392 of 605
Application +4 | -1
If tan⁡−1(1/4)+tan⁡−1(2/9)=(1/2)cos⁡−1x\tan ^{-1}(1 / 4)+\tan ^{-1}(2 / 9)=(1 / 2) \cos ^{-1} x then xx is equal to
View answer and explanation
Correct option: C
3/53 / 5
Option C (3/5) is correct because, after simplifying the left side using the tangent addition formula and then using the double angle formula for cosine, we find that x = 3/5.
393 Medium
393 of 605
Application +4 | -1
If cot⁡−1(nπ)>π6,n\cot ^{-1}\left(\frac{n}{\pi}\right)>\frac{\pi}{6}, n being a natural number, then maximum value of nn is
View answer and explanation
Correct option: B
5
Option B (5) is correct because solving the inequality cot⁡−1(nπ)>π6\cot^{-1}(\frac{n}{\pi}) > \frac{\pi}{6} leads to n<π3≈5.44n < \pi\sqrt{3} \approx 5.44, and since nn is a natural number, the maximum value is 5.
394 Easy
394 of 605
Remembering +4 | -1
Which of the following is the principal value branch of cosec⁡−1x\operatorname{cosec}^{-1} \mathrm{x} ?
View answer and explanation
Correct option: D
[−π2,π2]−{0}\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}
Option D ([−π2,π2]−{0}\left[\frac{{-\pi}}{{2}}, \frac{{\pi}}{{2}}\right] - \{0\}) is correct because the principal value branch of cosec⁡−1x\operatorname{cosec}^{-1} x is [−π2,π2]−{0}\left[\frac{{-\pi}}{{2}}, \frac{{\pi}}{{2}}\right] - \{0\}.
395 Easy
395 of 605
Analyzing +4 | -1
Let  SK=∑r=1Ktan⁡−1(6r22r+1+32r+1)\,{S_K} = \sum\limits_{r = 1}^K {{{\tan }^{ - 1}}\left( {\frac{{{6^r}}}{{{2^{2r + 1}} + {3^{2r + 1}}}}} \right)}. Then  lim⁡   K→∞ SK\,\mathop {\lim \,\,\,}\limits_{K \to \infty } \,{S_K} is equal to:
View answer and explanation
Correct option: B
cot⁡−1(32){\cot ^{ - 1}}\left( {\frac{3}{2}} \right)
Option B (cot⁡−1(32){\cot ^{ - 1}}\left( {\frac{3}{2}} \right)) is correct because the summation telescopes, and the limit is found using inverse trigonometric identities.
396 Medium
396 of 605
Application +4 | -1
tan⁡−1(1+33+3)+sec⁡−1(8+436+33)\tan ^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right)+\sec ^{-1}\left(\sqrt{\frac{8+4 \sqrt{3}}{6+3 \sqrt{3}}}\right) is equal to:
View answer and explanation
Correct option: A
π3\frac{\pi}{3}
The correct answer is A (π3\frac{{\pi}}{{3}}) because it is the sum of the simplified inverse trigonometric functions.
397 Easy
397 of 605
Analyzing +4 | -1
The number of roots of the equation sin⁡−1x+sin⁡−11−x3=π2\sin ^{-1} \sqrt{x}+\sin ^{-1} \sqrt{1-x^3}=\frac{\pi}{2} is equal to
View answer and explanation
Correct option: C
2
The correct answer is C (2) because solving the equation sin⁡−1x+sin⁡−11−x3=π2\sin^{-1} \sqrt{x} + \sin^{-1} \sqrt{1-x^3} = \frac{\pi}{2} yields two valid solutions: x = 0 and x = 1.
398 Easy
398 of 605
Application +4 | -1
If ∑r=120tan⁡−1(12r2)=tan⁡−1mn\sum_{\mathrm{r}=1}^{20} \tan ^{-1}\left(\frac{1}{2 \mathrm{r}^2}\right)=\tan ^{-1} \frac{\mathrm{m}}{\mathrm{n}}; where m\mathrm{m} and n\mathrm{n} are co-prime numbers then [log⁡2(m+n)]\left[\log _2(m+n)\right] equals (where [.] is greatest integer function)
View answer and explanation
Correct option: A
5
Option A (15) is correct because cot⁡71∘2=6+4+3+2\cot 7 \frac{1^{\circ}}{2} = \sqrt{6} + \sqrt{4} + \sqrt{3} + \sqrt{2}, and the sum of the numbers under the square roots is 15.
399 Easy
399 of 605
Application +4 | -1
The value of sin⁡−1(cos⁡33π5){\sin ^{ - 1}}\left( {\cos \frac{{33\pi }}{5}} \right) is
View answer and explanation
Correct option: D
−π10- \frac{\pi }{{10}}
Option D (−π10- \frac{\pi }{{10}}) is correct because simplifying the expression sin⁡−1(cos⁡33π5){\sin ^{ - 1}}\left( {\cos \frac{{33\pi }}{5}} \right) leads to −π10- \frac{\pi }{{10}}.
400 Easy
400 of 605
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 by using the identity sin⁡−1x+cos⁡−1x=π2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}, we can simplify the equation and solve for x.
401 Medium
401 of 605
Application +4 | -1
Solution of equation cos⁡−1x3+cos⁡−1x=π2\cos ^{-1} x \sqrt{3}+\cos ^{-1} x=\frac{\pi}{2} is
View answer and explanation
Correct option: A
x=12x=\frac{1}{2}
Option A (x=12x=\frac{1}{2}) is correct because it satisfies the given equation after simplification and considering the domain of inverse cosine function.
402 Easy
402 of 605
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.
403 Medium
403 of 605
Application +4 | -1
If cot⁡−1α+cot⁡−1β=cot⁡−1x\cot ^{-1} \alpha+\cot ^{-1} \beta=\cot ^{-1} x, then xx is equal to
View answer and explanation
Correct option: D
αβ−1α+β\frac{\alpha \beta-1}{\alpha+\beta}
Option D (αβ−1α+β\frac{{\alpha \beta-1}}{{\alpha+\beta}}) is correct because it is derived from the cotangent addition formula applied to the given equation.
404 Easy
404 of 605
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.
405 Medium
405 of 605
Understanding +4 | -1
Domain of f(x)=cos⁡−1x+cot⁡−1x+cosec⁡−1xf(x)=\cos ^{-1} x+\cot ^{-1} x+\operatorname{cosec}^{-1} x is
View answer and explanation
Correct option: D
{−1,1}\{-1,1\}
Option D ({−1,1-1,1}) is correct because the domain of cos⁡−1x\cos^{-1}x is [−1,1][-1, 1], the domain of cot⁡−1x\cot^{-1}x is R\mathbb{R}, and the domain of csc⁡−1x\csc^{-1}x is (−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty). The intersection of these domains is {−1,1-1, 1}.

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406 Easy
406 of 605
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].
407 Medium
407 of 605
Application +4 | -1
The value of [∣sin⁡(tan⁡−1(tan⁡3π4))∣]\left[\left|\sin \left(\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)\right)\right|\right] equals; where [.] is gint function and ∥\| is modulus function
View answer and explanation
Correct option: C
0
Option C (0) is correct because tan(3π/4) = -1, sin(tan⁻¹(-1)) = -1/√2, |sin(tan⁻¹(tan(3π/4)))| = 1/√2, and the greatest integer function [1/√2] = 0.
408 Easy
408 of 605
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.
409 Medium
409 of 605
Application +4 | -1
If x+1/x=5/2x+1 / x=5 / 2, then the principal value of sin⁡−1x\sin ^{-1} x is
View answer and explanation
Correct option: A
π/6\pi / 6
Option A, π/6\pi / 6, is correct because by solving the quadratic equation for sin⁡x\sin x, we find that sin⁡x=1/2\sin x = 1/2, which gives x=π/6x = \pi/6.
410 Medium
410 of 605
Analyzing +4 | -1
If 0<p<2π0 < p < 2 \pi and sin⁡−1(sin⁡p)<x2y+3y2x+181xy∀x,y∈R+\sin ^{-1}(\sin p) < \frac{x^2}{y}+\frac{3 y^2}{x}+\frac{1}{81 x y} \forall x, y \in R^{+}then pp cannot be equal to
View answer and explanation
Correct option: B
1.67
Option A (0) is correct because the equation has no solution since the LHS is always non-negative while the RHS is non-positive, and they cannot both be zero simultaneously.
411 Easy
411 of 605
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.
412 Easy
412 of 605
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.
413 Easy
413 of 605
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.
414 Hard
414 of 605
Application +4 | -1
If tan⁡−12x+tan⁡−13x=π4\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}, then xx is equal to
View answer and explanation
Correct option: B
16\frac{1}{6}
Option B (16\frac{1}{6}) is correct because solving the equation tan⁡−12x+tan⁡−13x=π4\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4} leads to x=16x = \frac{1}{6}, after discarding the extraneous solution.
415 Easy
415 of 605
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.
416 Medium
416 of 605
Application +4 | -1
If sin⁡−1(x)α=cos⁡−1xb=tan⁡−1yc,0<x<1,\frac{{{{\sin }^{ - 1}}\left( x \right)}}{\alpha } = \frac{{{{\cos }^{ - 1}}x}}{b} = \frac{{{{\tan }^{ - 1}}y}}{c},0 < x < 1, then the value of cos⁡(πca+b)\cos \left( {\frac{{\pi c}}{{a + b}}} \right) is
View answer and explanation
Correct option: C
1−y21+y2 \frac{{1 - {y^2}}}{{1 + {y^2}}}\,
Option C (1−y21+y2,\frac{{1 - {y^2}}}{{1 + {y^2}}}\\,) is correct because it is derived from trigonometric identities and the given relationships between inverse trigonometric functions.
417 Easy
417 of 605
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.
418 Easy
418 of 605
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.
419 Easy
419 of 605
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.
420 Medium
420 of 605
Application +4 | -1
The value of
sin⁡−1{cot⁡(sin⁡−12−34+cos⁡−1124+sec⁡−12)}\sin ^{-1}\left\{\cot \left(\sin ^{-1} \sqrt{\frac{2-\sqrt{3}}{4}}+\cos ^{-1} \frac{\sqrt{12}}{4}+\sec ^{-1} \sqrt{2}\right)\right\} is equal to
View answer and explanation
Correct option: C
0
Option C (0) is correct because the expression simplifies to sin⁡−1(cot⁡(π/2))\sin^{-1}(\cot(\pi/2)), which equals 0.
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