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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Mixed difficulty

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361 Easy
361 of 605
Application +4 | -1
The value of xx for which sin⁡[cot⁡−1(1+x)]=cos⁡(tan⁡−1x)\sin \left[ {{{\cot }^{ - 1}}\left( {1 + x} \right)} \right] = \cos \left( {{{\tan }^{ - 1}}x} \right) is
View answer and explanation
Correct option: D
−12- \frac{1}{2}
Option D (-1/2) is correct because it is the solution obtained by simplifying the given equation using trigonometric identities and solving for x.
362 Medium
362 of 605
Application +4 | -1
The value of {tan⁡[π4+12sin⁡−1(ab)]+tan⁡[π4−12sin⁡−1(ab)]}−1\left\{\tan \left[\frac{\pi}{4}+\frac{1}{2} \sin ^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\pi}{4}-\frac{1}{2} \sin ^{-1}\left(\frac{a}{b}\right)\right]\right\}^{-1}, where 0<a<b0 < a < b is
View answer and explanation
Correct option: C
b2−a22b\frac{\sqrt{b^2-a^2}}{2 b}
Option D (b2−a22b\frac{\sqrt{b^2-a^2}}{2 b}) is correct because simplifying the trigonometric expression and using the given condition leads to this result.
363 Medium
363 of 605
Application +4 | -1
If cosec⁡−1x=2cos⁡−17+cos⁡−1(35)\operatorname{cosec}^{-1} x=2 \cos ^{-1} 7+\cos ^{-1}\left(\frac{3}{5}\right) then the value of xx is
View answer and explanation
Correct option: B
125/117125 / 117
Option B (125117\frac{{125}}{{117}}) is correct because it correctly uses the cotangent addition formula and trigonometric identities to find the value of x.
364 Medium
364 of 605
Analyzing +4 | -1
cos⁡−1{12x2+1−x21−x24}=cos⁡−1x2−cos⁡−1x{\cos ^{ - 1}}\left\{ {\frac{1}{2}{x^2} + \sqrt {1 - {x^2}} \sqrt {1 - \frac{{{x^2}}}{4}} } \right\} = {\cos ^{ - 1}}\frac{x}{2} - {\cos ^{ - 1}}x holds for
View answer and explanation
Correct option: A
∣x∣≤1\left| x \right| \le 1
Option A (∣x∣≤1\left| x \right| \le 1) is correct because the domain of the inverse cosine function requires ∣x∣≤1|x| \le 1, and this condition must be satisfied for all terms in the equation.
365 Easy
365 of 605
Application +4 | -1
The value of sin⁡−1(cos⁡2)−cos⁡−1(sin⁡2)+tan⁡−1(cot⁡4)−cot⁡−1(tan⁡4)\sin ^{-1}(\cos 2)-\cos ^{-1}(\sin 2)+\tan ^{-1}(\cot 4)-\cot ^{-1}(\tan 4) +sec⁡−1(cosec⁡6)−cosec⁡−1(sec⁡6)+\sec ^{-1}(\operatorname{cosec} 6)-\operatorname{cosec}^{-1}(\sec 6) is
View answer and explanation
Correct option: D
5π−165 \pi-16
Option D (5π−165 \pi-16) is correct because it is the simplified value of the given expression involving inverse trigonometric functions.
366 Easy
366 of 605
Application +4 | -1
Value of sin⁡(sin⁡−1(3/5)+cos⁡−1(4/5))\sin \left(\sin ^{-1}(3 / 5)+\cos ^{-1}(4 / 5)\right) is equal to
View answer and explanation
Correct option: B
24/2524 / 25
Option B (24/2524/25) is correct because it results from applying the sine addition formula and simplifying the expression.
367 Easy
367 of 605
Application +4 | -1
If α=cos⁡−1(35),β=tan⁡−1(13)\alpha=\cos ^{-1}\left(\frac{3}{5}\right), \beta=\tan ^{-1}\left(\frac{1}{3}\right), where 0<α,β<π20<\alpha, \beta<\frac{\pi}{2}, then α−β\alpha-\beta is equal to :
View answer and explanation
Correct option: D
sin⁡−1(9510)\sin ^{-1}\left(\frac{9}{5 \sqrt{10}}\right)
The correct answer is D (sin⁡−1(9510)\sin^{-1}(\frac{9}{5 \sqrt{10}})) because it is obtained by using the tangent subtraction formula and then converting to arcsin.
368 Easy
368 of 605
Application +4 | -1
If x=2/3x=2 / 3, then sin⁡2(tan⁡−1x)+cos⁡2(sin⁡−1x)\sin ^2\left(\tan ^{-1} x\right)+\cos ^2\left(\sin ^{-1} x\right) is equal to
View answer and explanation
Correct option: C
101/117101 / 117
Option C (101/117101 / 117) is correct because by substituting x=2/3x=2/3 and simplifying the expression, we get the answer 101/117101/117.
369 Medium
369 of 605
Application +4 | -1
Considering only the principal values of inverse trigonometric functions, the number of positive real values of xx satisfying tan⁡−1(x)+tan⁡−1(2x)=π4\tan ^{-1}(x)+\tan ^{-1}(2 x)=\frac{\pi}{4} is:
View answer and explanation
Correct option: B
1
Option B (1) is correct because the function is monotonically increasing, ensuring only one positive real solution for the given equation.
370 Easy
370 of 605
Application +4 | -1
If 0<x<120 < x < \frac{1}{{\sqrt 2 }} and sin⁡−1xα=cos⁡−1xβ\frac{{{{\sin }^{ - 1}}x}}{\alpha } = \frac{{{{\cos }^{ - 1}}x}}{\beta } then a value of sin⁡(2παα+β)\sin \left( {\frac{{2\pi \alpha }}{{\alpha + \beta }}} \right) is
View answer and explanation
Correct option: B
4x(1−x2)(1−2x2)4 x \sqrt{\left(1-x^2\right)}\left(1-2 x^2\right)
Option B is correct because by using trigonometric identities and the given condition, we can simplify the expression and find the value of sin⁡(2παα+β)\sin \left( {\frac{{2\pi \alpha }}{{\alpha + \beta }}} \right).
371 Medium
371 of 605
Application +4 | -1
If x1=2tan⁡−1(1+x1−x),x2=sin⁡−1(1−x21+x2)x_1=2 \tan ^{-1}\left(\frac{1+x}{1-x}\right), x_2=\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right); where x∈x \in (0,1)(0,1) then x1+x2x_1+x_2 is equal to
View answer and explanation
Correct option: C
π\pi
Option C (π\pi) is correct because simplifying the expression using trigonometric identities and the given condition leads to π\pi.
372 Medium
372 of 605
Analyzing +4 | -1
Sum of series of cot⁡−153+cot⁡−193+cot⁡−1153+cot⁡−1233+……∞\cot ^{-1} \frac{5}{\sqrt{3}}+\cot ^{-1} \frac{9}{\sqrt{3}}+\cot ^{-1} \frac{15}{\sqrt{3}}+\cot ^{-1} \frac{23}{\sqrt{3}}+\ldots \ldots \infty is equal to
View answer and explanation
Correct option: B
π3\frac{\pi}{3}
Option B (π3\frac{\pi}{3}) is correct because the sum of the infinite series of inverse cotangents telescopes to π3\frac{\pi}{3} after converting to inverse tangents and simplifying.
373 Easy
373 of 605
Application +4 | -1
4 tan⁡−115−tan⁡−11239  {\tan ^{ - 1}}\frac{1}{5} - {\tan ^{ - 1}}\frac{1}{{239}}\; is equal to
View answer and explanation
Correct option: D
π4\frac{\pi }{4}
Option D (π4\frac{\pi }{4}) is correct because by using the double angle formula for the inverse tangent function and then the tangent subtraction formula, we can simplify the expression to tan⁡−11{\tan ^{ - 1}}1, which equals π4\frac{\pi }{4}.
374 Easy
374 of 605
Application +4 | -1
The principal value of tan⁡−1(cot⁡43π4)\tan ^{-1}\left(\cot \frac{43 \pi}{4}\right) is:
View answer and explanation
Correct option: C
−π4-\frac{\pi}{4}
The correct answer is C (−π4\frac{-\pi}{4}) because it involves simplifying the cotangent and using the properties of inverse trigonometric functions.
375 Medium
375 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+\tan ^{-1}(x / y)+\tan ^{-1}[(x+ y)/(x−y)]y) /(x-y)] is equal to:
View answer and explanation
Correct option: C
3π/43 \pi / 4
Option C (3π4\frac{{3 \pi}}{{4}}) is correct because the given expression simplifies to π+tan⁡−1(−1)\pi + \tan^{{-1}}(-1) under the given conditions, which equals 3π4\frac{{3\pi}}{{4}}.

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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
376 Easy
376 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 is correct because the expression simplifies using trigonometric identities and inverse trigonometric functions to x2+1x2+2\sqrt {\frac{{{x^2} + 1}}{{{x^2} + 2}}}.
377 Easy
377 of 605
Application +4 | -1
If x=sin⁡(2tan⁡−12)x = \sin \left( {2{{\tan }^{ - 1}}2} \right) and y=sin⁡(12tan⁡−143)y = \sin \left( {\frac{1}{2}{{\tan }^{ - 1}}\frac{4}{3}} \right), then
View answer and explanation
Correct option: B
y2=1−x{y^2} = 1 - x
Option B (y2=1−x{y^2} = 1 - x) is correct because it represents the relationship between x and y after simplifying the given trigonometric expressions.
378 Easy
378 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 tangent subtraction formula, we can simplify the expression and find that A - B equals 30 degrees.
379 Medium
379 of 605
Application +4 | -1
The value of the expression sec⁡2(tan⁡−12)+cosec⁡2(cot⁡−13)+cosec⁡(2cot⁡−12+cos⁡−135)\sec ^2\left(\tan ^{-1} 2\right)+\operatorname{cosec}^2\left(\cot ^{-1} 3\right)+\operatorname{cosec}\left(2 \cot ^{-1} 2+\cos ^{-1} \frac{3}{5}\right) is equal to:
View answer and explanation
Correct option: D
15+252415+\frac{25}{24}
Option D (15+252415+\frac{{25}}{{24}}) is correct because simplifying the expression using trigonometric identities and inverse trigonometric functions results in 15+252415+\frac{{25}}{{24}}.
380 Easy
380 of 605
Analyzing +4 | -1
2π−(sin⁡−145+sin⁡−1513+sin⁡−11665)2 \pi-\left(\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{16}{65}\right) is equal to :
View answer and explanation
Correct option: C
3π2\frac{3 \pi}{2}
Option C (3π2\frac{{3 \pi}}{{2}}) is correct because by converting the inverse sines to inverse tangents and using the tangent addition formula, the expression simplifies to 2π−π2=3π22\pi - \frac{\pi}{2} = \frac{{3\pi}}{{2}}.
381 Easy
381 of 605
Application +4 | -1
Sin⁡(2sin⁡−10.8)=\operatorname{Sin}\left(2 \sin ^{-1} 0.8\right)=
View answer and explanation
Correct option: A
0.96
Option A (0.96) is correct because it is the result of applying the double angle formula for sine and simplifying the expression.
382 Medium
382 of 605
Application +4 | -1
The solution of the inequality (cot⁡−1x)2−5cot⁡−1x+6>0\left(\cot ^{-1} x\right)^2-5 \cot ^{-1} x+6>0 is
View answer and explanation
Correct option: B
(−∞,cot⁡3)∪(cot⁡2,∞)(-\infty, \cot 3) \cup(\cot 2, \infty)
Option B ((−∞,cot⁡3)∪(cot⁡2,∞)(-\infty, \cot 3) \cup (\cot 2, \infty)) is correct because solving the inequality for cot⁡−1x\cot^{-1}x and then finding the corresponding values of xx gives the solution set.
383 Easy
383 of 605
Application +4 | -1
tan⁡−1xa2−x2\tan ^{-1} \frac{x}{\sqrt{a^2-x^2}} is equal to
View answer and explanation
Correct option: C
sin⁡−1(xa)\sin ^{-1}\left(\frac{x}{a}\right)
Option C (sin⁡−1(xa)\sin ^{-1}(\frac{{x}}{{a}})) is correct because by substituting x=asin⁡θx = a \sin \theta, the expression simplifies to this inverse sine function.
384 Easy
384 of 605
Analyzing +4 | -1
If xx takes negative permissible value, then sin⁡−1x{\sin ^{ - 1}}x is equal to
View answer and explanation
Correct option: A
−cos⁡−11−x2- {\cos ^{ - 1}}\sqrt {1 - {x^2}}
Option A is correct because it uses the property that sin-1(-x) = -sin-1(x) and the relationship between sin-1(x) and cos-1(sqrt(1-x2)) when x is negative.
385 Easy
385 of 605
Application +4 | -1
cot⁡−1[(cos⁡α)1/2]−tan⁡−1[(cos⁡α)1/2]=x\cot ^{-1}\left[(\cos \alpha)^{1 / 2}\right]-\tan ^{-1}\left[(\cos \alpha)^{1 / 2}\right]=x then sin⁡x=\sin x=
View answer and explanation
Correct option: A
tan⁡2(α/2)\tan ^2(\alpha / 2)
Option A (tan⁡2(α/2)\tan ^2(\alpha / 2)) is correct because by simplifying the given equation using trigonometric identities, we find that sin⁡x=tan⁡2(α/2)\sin x = \tan^2(\alpha / 2).
386 Medium
386 of 605
Application +4 | -1
Sum to nn terms of the series
cosec⁡−110+cosec⁡−150+cosec⁡−1170+……+cosec⁡−1(n2+1)(n2+2n+2) is \operatorname{cosec}^{-1} \sqrt{10}+\operatorname{cosec}^{-1} \sqrt{50}+\operatorname{cosec}^{-1} \sqrt{170}+\ldots \ldots \\ +\operatorname{cosec}^{-1} \sqrt{\left(n^2+1\right)\left(n^2+2 n+2\right)} \text { is }
View answer and explanation
Correct option: A
tan⁡−1(n+1)−π4\tan ^{-1}(n+1)-\frac{\pi}{4}
Option A is correct because the sum of the series can be simplified using the arctangent subtraction formula, resulting in tan⁡−1(n+1)−π4\tan^{-1}(n+1) - \frac{\pi}{4}.
387 Easy
387 of 605
Application +4 | -1
If x2+y2+z2=r2,{x^2} + {y^2} + {z^2} = {r^2},, then

tan⁡−1(xyzr)+tan⁡−1(yzxr)+tan⁡−1(xzyr)  {\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{xz}}{{yr}}} \right)\; is equal to
View answer and explanation
Correct option: B
π2\frac{\pi }{2}
Option B (π2\frac{\pi}{2}) is correct because the expression simplifies to tan⁡−1(∞)\tan^{-1}(\infty) using the given condition x2+y2+z2=r2{x^2} + {y^2} + {z^2} = {r^2}.
388 Easy
388 of 605
Application +4 | -1
tan⁡−1(2x1−x2)+cot⁡−1(1−x22x)=π3,x∈[−1,1]\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{3}, x \in[-1,1] sum of all solutions is α−43\alpha-\frac{4}{\sqrt{3}}, then α\alpha is:
View answer and explanation
Correct option: B
2
Option B (2) is correct because after solving the equation, the sum of all solutions is indeed 2 - 4/sqrt(3), making alpha equal to 2.
389 Easy
389 of 605
Application +4 | -1
The value of sin⁡[π2−sin⁡−1  (−32)]\sin \left[ {\frac{\pi }{2} - {{\sin }^{ - 1}}\;\left( { - \frac{{\sqrt 3 }}{2}} \right)} \right] is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (1/2) is correct because simplifying the expression using trigonometric identities and inverse trigonometric functions leads to this result.
390 Easy
390 of 605
Application +4 | -1
If tan⁡−111+2+tan⁡−111+(2)(3)+tan⁡−111+(3)(4)+…tan⁡−111+n(n+1)=tan⁡−1θ\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, then θ=\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 series telescopes to tan⁡−1(n+1)−tan⁡−1(1)\tan^{-1}(n+1) - \tan^{-1}(1), which simplifies to tan⁡−1(nn+2)\tan^{-1}(\frac{n}{n+2}).
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