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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Medium

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Showing 31-60 of 221 questions
31 Medium
31 of 221
Application +4 | -1
The value of sin⁡−1(sin⁡10)\sin ^{-1}(\sin 10) is -
View answer and explanation
Correct option: C
3π−103 \pi-10
Option C (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 we need to find an angle in this range that has the same sine value as 10.
32 Medium
32 of 221
Analyzing +4 | -1
If (x−1)(x2+2)>0(x-1)\left(x^2+2\right) > 0, then the value of sin⁡(12tan⁡−1(2x1−x2)−tan⁡−1x)\sin \left(\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^2}\right)-\tan ^{-1} x\right) is
View answer and explanation
Correct option: B
−1-1
Option B (-1) is correct because, given the condition x > 1, the expression simplifies to sin(-π/2), which equals -1.
33 Medium
33 of 221
Application +4 | -1
If cos⁡−1p+cos⁡−11−p+cos⁡−11−q=3π4\cos ^{-1} \sqrt{p}+\cos ^{-1} \sqrt{1-p}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}, then the value of qq is
View answer and explanation
Correct option: A
0.5
Option A (0.5) is correct because by using trigonometric identities and solving the equation, the value of q is found to be 1/2.
34 Medium
34 of 221
Application +4 | -1
cosec⁡−1(cos⁡x)\operatorname{cosec}^{-1}(\cos x) is real if
View answer and explanation
Correct option: D
xx is integer multiple of π\pi
Option D (xx is integer multiple of π\pi) is correct because for cosec⁡−1(cos⁡x)\operatorname{cosec}^{-1}(\cos x) to be real, ∣cos⁡x∣|\cos x| must be equal to 1, which occurs when xx is an integer multiple of π\pi.
35 Medium
35 of 221
Application +4 | -1
The value of tan⁡(tan⁡−11+tan⁡−12+tan⁡−13)=\tan \left(\tan ^{-1} 1+\tan ^{-1} 2+\tan ^{-1} 3\right)=
View answer and explanation
Correct option: A
0
Option A (0) is correct because the sum of the inverse tangents simplifies to 0.
36 Medium
36 of 221
Application +4 | -1
If sin⁡−1x+sin⁡−1y+sin⁡−1z=π\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\pi, then x1−x2+y1−y2+z1−z2x \sqrt{1-x^2}+y \sqrt{1-y^2}+z \sqrt{1-z^2} equals
View answer and explanation
Correct option: D
2xyz2 x y z
Option D (2xyz2xyz) is correct because, given the condition, the expression simplifies to 2xyz2xyz using trigonometric identities.
37 Medium
37 of 221
Application +4 | -1
If 12sin⁡−1(3sin⁡2θ5+4cos⁡2θ)=π4\frac{1}{2} \sin ^{-1}\left(\frac{3 \sin 2 \theta}{5+4 \cos 2 \theta}\right)=\frac{\pi}{4}, then tan⁡θ\tan \theta is equal to
View answer and explanation
Correct option: B
3
Option B (3) is correct because solving the equation after applying the sine function and simplifying leads to tan⁡θ=3\tan \theta = 3.
38 Medium
38 of 221
Analyzing +4 | -1
sin⁡−112+sin⁡−12−16+sin⁡−13−212+sin⁡−12−320+sin⁡−15−230+…∞\sin ^{-1} \frac{1}{\sqrt{2}}+\sin ^{-1} \frac{\sqrt{2}-1}{\sqrt{6}}+\sin ^{-1} \frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+\sin ^{-1} \frac{2-\sqrt{3}}{\sqrt{20}}+\sin ^{-1} \frac{\sqrt{5}-2}{\sqrt{30}}+\ldots \infty
View answer and explanation
Correct option: B
π2\frac{\pi}{2}
The correct answer is π2\frac{\pi}{2} because the given series telescopes to π2\frac{\pi}{2} as nn approaches infinity.
39 Medium
39 of 221
Application +4 | -1
cot⁡−1xy+1x−y+cot⁡−1yz+1y−z+cot⁡−1zx+1z−x;x>y>z>0\cot ^{-1} \frac{x y+1}{x-y}+\cot ^{-1} \frac{y z+1}{y-z}+\cot ^{-1} \frac{z x+1}{z-x} ; x > y > z > 0 and x,y,zx, y, z distinct is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because the expression simplifies to 0 using the inverse tangent subtraction formula and the given conditions.
40 Medium
40 of 221
Application +4 | -1
2tan⁡−1(cos⁡x)=tan⁡−1(cosec⁡2x)2 \tan ^{-1}(\cos x)=\tan ^{-1}\left(\operatorname{cosec}^2 x\right), then x=x=
View answer and explanation
Correct option: D
π3\frac{\pi}{3}
Option D (π3\frac{\pi}{3}) is correct because solving the trigonometric equation 2tan⁡−1(cos⁡x)=tan⁡−1(cosec⁡2x)2 \tan^{-1}(\cos x) = \tan^{-1}(\operatorname{cosec}^2 x) leads to x=π3x = \frac{\pi}{3}.
41 Medium
41 of 221
Analyzing +4 | -1
Let SS be solution set for values of xx satisfying cos⁡−1(2x)+cos⁡−11−x2=π\cos ^{-1}(2 x)+\cos ^{-1} \sqrt{1-x^2}=\pi, then ∑x∈S2sin⁡−1(x2−1)\sum\limits_{{x \in S}} 2 \sin ^{-1}\left(x^2-1\right) is equal to:
View answer and explanation
Correct option: B
−sin⁡−1(2425)-\sin ^{-1}\left(\frac{24}{25}\right)
Option B (−sin⁡−1(2425)-\sin ^{-1}\left(\frac{24}{25}\right)) is correct because solving the equation for xx and substituting it back into the expression yields −sin⁡−1(2425)-\sin ^{-1}\left(\frac{24}{25}\right).
42 Medium
42 of 221
Application +4 | -1
If cos⁡−1x−sin⁡−1x=0\cos ^{-1} x-\sin ^{-1} x=0, then xx is equal to
View answer and explanation
Correct option: C
12\frac{1}{\sqrt{2}}
Option C (12\frac{{1}}{{\sqrt{{2}}}}) is correct because it is the value of x that satisfies the equation cos⁡−1x−sin⁡−1x=0\cos^{-1} x - \sin^{-1} x = 0.
43 Medium
43 of 221
Application +4 | -1
cos⁡[2cos⁡−115+sin⁡−115]\cos \left[2 \cos ^{-1} \frac{1}{5}+\sin ^{-1} \frac{1}{5}\right] is equal to
View answer and explanation
Correct option: B
−265-\frac{2 \sqrt{6}}{5}
Option B (−265-\frac{2 \sqrt{6}}{5}) is correct because using the identity sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} and then simplifying the cosine expression leads to the correct answer.
44 Medium
44 of 221
Application +4 | -1
The smallest and the largest values of tan⁡−1(1−x1+x),  0≤x≤1{\tan ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right),\;0 \le x \le 1 are
View answer and explanation
Correct option: B
0,π40,\frac{\pi }{4}
Option B (0,π40,\frac{\pi }{4}) is correct because it represents the range of the given function when xx varies from 0 to 1.
45 Medium
45 of 221
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 the correct condition for the given trigonometric equation to be possible, ensuring the arguments of the inverse trigonometric functions are within the valid range.

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46 Medium
46 of 221
Application +4 | -1
If A=tan⁡−117,B=tan⁡−113A=\tan ^{-1} \frac{1}{7}, B=\tan ^{-1} \frac{1}{3}, then
View answer and explanation
Correct option: D
cos⁡2A=sin⁡4B\cos 2 A=\sin 4 B
Option D (cos⁡2A=sin⁡4B\cos 2 A=\sin 4 B) is correct because, given the values of A and B, it can be shown that cos⁡2A\cos 2A and sin⁡4B\sin 4B both equal 2425\frac{{24}}{{25}}.
47 Medium
47 of 221
Analyzing +4 | -1
If x,  y,  zx,\;y,\;z are in AP and tan⁡−1x,tan⁡−1y{\tan ^{ - 1}}x,{\tan ^{ - 1}}y and tan⁡−1z{\tan ^{ - 1}}z are also in AP, then
View answer and explanation
Correct option: A
x=y=zx = y = z
Option A is correct because if x, y, and z are in AP and their inverse tangents are also in AP, then the only solution is when x = y = z.
48 Medium
48 of 221
Application +4 | -1
If f(x)=ecosecos⁡−1x2+tan⁡−1cot⁡−1x2f(x)=\mathrm{e}^{\operatorname{cosecos}^{-1} x^2+\tan ^{-1} \cot ^{-1} x^2} then minimum value of f(x)f(x) is
View answer and explanation
Correct option: B
e2e^2
Option B (e2e^2) is correct because the minimum value of the function is obtained when x2+1x2x^2 + \frac{1}{x^2} is minimized, which occurs at x2=1x^2 = 1.
49 Medium
49 of 221
Analyzing +4 | -1
If y=tan⁡−1(log⁡e(e/x2)log⁡e(ex2))+tan⁡−1(3+2log⁡ex1−6log⁡ex),then  d2ydx2  isy = {\tan ^{ - 1}}\left( {\frac{{{{\log }_e}\left( {e/{x^2}} \right)}}{{{{\log }_e}\left( {e{x^2}} \right)}}} \right) + {\tan ^{ - 1}}\left( {\frac{{3 + 2{{\log }_e}x}}{{1 - 6{{\log }_e}x}}} \right),{\text{then}}\,\,\frac{{{d^2}y}}{{d{x^2}}}\,\,{\text{is}}
View answer and explanation
Correct option: C
0
Option D ({0}) is correct because the expression inside the square root must be non-negative, which implies that {x} must be 0.
50 Medium
50 of 221
Application +4 | -1
If cot⁡−1(1a)+cot⁡−1(1b)+cot⁡−1(1c)=π2\cot ^{-1}\left(\frac{1}{a}\right)+\cot ^{-1}\left(\frac{1}{b}\right)+\cot ^{-1}\left(\frac{1}{c}\right)=\frac{\pi}{2}, then
View answer and explanation
Correct option: B
ab+bc+ca=1a b+b c+c a=1
Option B (ab+bc+ca=1a b+b c+c a=1) is correct because it is derived from the tangent addition formula and the given condition.
51 Medium
51 of 221
Analyzing +4 | -1
If a,  ba,\;b are positive quantities and if a1=a+b2,  b1=a1b,  a2=a1+b12,  b2=a2b1{a_1} = \frac{{a + b}}{2},\;{b_1} = \sqrt {{a_1}b} ,\;{a_2} = \frac{{{a_1} + {b_1}}}{2},\;{b_2} = \sqrt {{a_2}{b_1}} and so on, then
View answer and explanation
Correct option: B
b∞=b2−a2cos⁡−1(ab){b_\infty } = \frac{{\sqrt {{b^2} - {a^2}} }}{{{{\cos }^{ - 1}}\left( {\frac{a}{b}} \right)}}
Option D is incorrect. The correct answer is b∞=b2−a2cos⁡−1(ab){b_\infty } = \frac{{\sqrt {{{{b^2} - {a^2}}}} }}{{{{\cos }^{ - 1}}\left( {\frac{a}{b}} \right)}} which is not listed, and the provided solution derives this result using trigonometric substitution and limit evaluation.
52 Medium
52 of 221
Application +4 | -1
If in a Δ  ABC,∠A=tan⁡−12  and  ∠B=tan⁡−13,  \Delta \;ABC,\angle A = {\tan ^{ - 1}}2\;\mathrm{and}\;\angle B = {\tan ^{ - 1}}3,\; then angle CC is equal to
View answer and explanation
Correct option: C
π4\frac{\pi }{4}
Option C (π4\frac{\pi }{4}) is correct because the sum of angles A and B is 3π\pi/4, making angle C equal to π\pi - 3π\pi/4 = π\pi/4.
53 Medium
53 of 221
Application +4 | -1
cos⁡−1[cos⁡(−17π5)]\cos ^{-1}\left[\cos \left(\frac{-17 \pi}{5}\right)\right] is equal to
View answer and explanation
Correct option: B
3π5\frac{3 \pi}{5}
Option B (3π5\frac{{3 \pi}}{{5}}) is correct because by applying the properties of cosine and inverse cosine functions, the expression simplifies to 3π5\frac{{3\pi}}{{5}}.
54 Medium
54 of 221
Application +4 | -1
If (sin⁡−1x)2+(sin⁡−1y)2+2sin⁡−1xsin⁡−1y=π2\left(\sin ^{-1} x\right)^2+\left(\sin ^{-1} y\right)^2+2 \sin ^{-1} x \sin ^{-1} y=\pi^2, then x2+y2x^2+y^2 is equal to:
View answer and explanation
Correct option: C
2
Option C (2) is correct because the given equation can be simplified to find the values of x and y, and then x2+y2x^2 + y^2 can be calculated.
55 Medium
55 of 221
Application +4 | -1
The number of positive integral pairs (a,b)(a, b) satisfying the equation tan⁡−1a+tan⁡−1b=tan⁡−17\tan ^{-1} a+\tan ^{-1} b=\tan ^{-1} 7 is
View answer and explanation
Correct option: B
2
Option B, 2, is correct because by applying the tangent addition formula and solving for a in terms of b, we find only two positive integer pairs (a, b) that satisfy the equation.
56 Medium
56 of 221
Application +4 | -1
The value of sin⁡−1[cos⁡(tan⁡−113+tan⁡−112)]\sin ^{-1}\left[\cos \left(\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{2}\right)\right] is π/k\pi / k; then find the value of kk.
View answer and explanation
Correct option: A
4
Option A (4) is correct because the expression simplifies to sin⁡−1(cos⁡(π4))=sin⁡−1(22)=π4\sin^{-1}(\cos(\frac{\pi}{4})) = \sin^{-1}(\frac{\sqrt{2}}{2}) = \frac{\pi}{4}, thus k=4.
57 Medium
57 of 221
Application +4 | -1
If tan⁡−12p+tan⁡−13p+tan⁡−14p+tan⁡−15p=π2\tan ^{-1} \frac{2}{p}+\tan ^{-1} \frac{3}{p}+\tan ^{-1} \frac{4}{p}+\tan ^{-1} \frac{5}{p}=\frac{\pi}{2}, then (p2+120p2)\left(p^2+\frac{120}{p^2}\right) is equal to
View answer and explanation
Correct option: B
71
Option B (71) is correct because solving the given equation involving inverse tangents leads to the equation p4+120=71p2p^4 + 120 = 71p^2, which directly gives p2+120p2=71p^2 + \frac{120}{p^2} = 71.
58 Medium
58 of 221
Application +4 | -1
Iftan⁡−111+2+tan⁡−111+(2)(3)+tan⁡−111+(3)(4)+…+tan⁡−111+n(n+1)=tan⁡−1θ then θ=\text{If} \tan ^{-1} \frac{1}{1+2}+\tan ^{-1} \frac{1}{1+(2)(3)}+\tan ^{-1} \frac{1}{1+(3)(4)}+\ldots+\tan ^{-1} \frac{1}{1+n(n+1)}=\tan ^{-1}\theta ~\text{then} ~\theta=
View answer and explanation
Correct option: C
nn+2\frac{n}{n+2}
Option C (nn+2\frac{n}{n+2}) is correct because the given sum telescopes, simplifying to an expression whose arctangent is easily calculated.
59 Medium
59 of 221
Application +4 | -1
The value of cot⁡[∑n=123cot⁡−1(1+∑k=1n2k)]\cot \left[\sum\limits_{{n=1}}^{23} \cot ^{-1}\left(1+\sum\limits_{{k=1}}^n 2 k\right)\right] is
View answer and explanation
Correct option: B
2523\frac{25}{23}
Option B (25/23) is correct because the summation of the inverse cotangent series simplifies to a single inverse cotangent, and taking the cotangent of that result yields 25/23.
60 Medium
60 of 221
Application +4 | -1
tan⁡(π4+12cos⁡−1x)+tan⁡(π4−12cos⁡−1x),x≠0\tan \left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x\right)+\tan \left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\right), x \neq 0, is equal to
View answer and explanation
Correct option: C
2x\frac{2}{x}
Option C (2x\frac{2}{x}) is correct because the given expression simplifies to 2x\frac{2}{x} using trigonometric identities and substitution.
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