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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Easy

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Showing 301-330 of 365 questions

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301 Easy
301 of 365
Application +4 | -1
If x+y+z=xyzx + y + z = xyz, then tan⁡−1x+tan⁡−1y+tan⁡−1z={\tan ^{ - 1}}x + {\tan ^{ - 1}}y + {\tan ^{ - 1}}z =
View answer and explanation
Correct option: A
0
Option D (None of these) is correct because the given solution incorrectly concludes that the expression equals 0. The correct simplification leads to a different result.
302 Easy
302 of 365
Application +4 | -1
The solutions of the equation tan⁡−1(x+1)+tan⁡−1(x−1)=tan⁡−1831{\tan ^{ - 1}}\left( {x + 1} \right) + {\tan ^{ - 1}}\left( {x - 1} \right) = {\tan ^{ - 1}}\frac{8}{{31}} are
View answer and explanation
Correct option: B
14,  −8\frac{1}{4},\; - 8
Option B (14,−8\frac{1}{4}, -8) is correct because these are the solutions obtained after solving the given inverse trigonometric equation using the tangent addition formula.
303 Easy
303 of 365
Application +4 | -1
If 4sin⁡−1x+cos⁡−1x=π,    4{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \pi ,\;\; then xx is equal to
View answer and explanation
Correct option: B
1/21/2
Option B (1/21/2) is correct because solving the given equation 4sin⁡−1x+cos⁡−1x=π4{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \pi leads to x=1/2x = 1/2.
304 Easy
304 of 365
Application +4 | -1
If cos⁡−1x+cos⁡−1y+cos⁡−1z=π,{\cos ^{ - 1}}x + {\cos ^{ - 1}}y + {\cos ^{ - 1}}z = \pi , then
View answer and explanation
Correct option: C
x2+y2+z2=1−2xyz{x^2} + {y^2} + {z^2} = 1 - 2xyz
Option C (x2+y2+z2=1−2xyz{x^2} + {y^2} + {z^2} = 1 - 2xyz) is correct because it is the simplified form of the given trigonometric equation after applying inverse trigonometric identities and algebraic manipulations.
305 Easy
305 of 365
Application +4 | -1
If the (cos⁡−1x)=sin⁡(cot⁡−112),  ({\cos ^{ - 1}}x) = \sin \left( {{{\cot }^{ - 1}}\frac{1}{2}} \right),\; then xx is equal to
View answer and explanation
Correct option: B
±53\pm \frac{{\sqrt 5 }}{3}
The correct answer is B (±53\pm \frac{{\sqrt 5 }}{3}) because by using trigonometric identities and simplification, we can find the value of x.
306 Easy
306 of 365
Application +4 | -1
If cos⁡−1x+cos⁡−1y=π/2\cos ^{-1} x+\cos ^{-1} y=\pi / 2 and tan⁡−1x−tan⁡−1y=0\tan ^{-1} x-\tan ^{-1} y=0 then x2+xy+y2x^2+x y+y^2 is equal to
View answer and explanation
Correct option: C
3/23 / 2
Option C, 3/23/2, is correct because by using the given conditions, we find that x=y=1/2x = y = 1/\sqrt{2}, and substituting these values into the expression x2+xy+y2x^2 + xy + y^2 gives 3/23/2.
307 Easy
307 of 365
Application +4 | -1
If cos⁡−135−sin⁡−145=cos⁡−1x,{\cos ^{ - 1}}\frac{3}{5} - {\sin ^{ - 1}}\frac{4}{5} = {\cos ^{ - 1}}x, then xx is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because simplifying the equation using trigonometric identities leads to cos⁡−1x=0\cos^{-1}x = 0, which implies x = 1.
308 Easy
308 of 365
Analyzing +4 | -1
The sum of the infinite series

sin⁡−1(12)+sin⁡−1(2−16)+sin⁡−1(3−212)+...{\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) + {\sin ^{ - 1}}\left( {\frac{{\sqrt 2 - 1}}{{\sqrt 6 }}} \right) + {\sin ^{ - 1}}\left( {\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt {12} }}} \right) + ...

+...+sin⁡−1(n−(n−1){n(n+1)})+...+ ... + {\sin ^{ - 1}}\left( {\frac{{\sqrt n - \sqrt {\left( {n - 1} \right)} }}{{\sqrt {\left\{ {n\left( {n + 1} \right)} \right\}} }}} \right) + ... is
View answer and explanation
Correct option: C
π2\frac{\pi }{2}
Option C (π2\frac{\pi}{2}) is correct because the series telescopes to tan⁡−1(n){\tan^{-1}}(\sqrt{n}) which approaches π2\frac{\pi}{2} as n approaches infinity.
309 Easy
309 of 365
Application +4 | -1
The value of sin (cot⁡−1x)({\cot ^{ - 1}}x) is
View answer and explanation
Correct option: D
(1+x2)−1/2{\left( {1 + {x^2}} \right)^{ - 1/2}}
Option D ((1+x2)−1/2{\left( {1 + {x^2}} \right)^{ - 1/2}}) is correct because by using trigonometric identities and the definition of cotangent, we can express sin⁡(cot⁡−1x)\sin({\cot ^{ - 1}}x) as (1+x2)−1/2{\left( {1 + {x^2}} \right)^{ - 1/2}}.
310 Easy
310 of 365
Application +4 | -1
If tan⁡−1x+tan⁡−1y+tan⁡−1z=π\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\pi, then the value of x+y+z−xyzx+y+z-x y z is
View answer and explanation
Correct option: A
0
Option A (0) is correct because if tan⁡−1x+tan⁡−1y+tan⁡−1z=π\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi, then x+y+z=xyzx + y + z = xyz, which implies x+y+z−xyz=0x + y + z - xyz = 0.
311 Easy
311 of 365
Application +4 | -1
If (tan⁡−1x)2+(cot⁡−1x)2=5π28{\left( {{{\tan }^{ - 1}}x} \right)^2} + {\left( {{{\cot }^{ - 1}}x} \right)^2} = \frac{{5{\pi ^2}}}{8}, then xx is equal to
View answer and explanation
Correct option: D
−1- 1
Option D (-1) is correct because solving the equation using the property (tan-1x)2 + (cot-1x)2 = 5pi2/8 leads to x = -1.
312 Easy
312 of 365
Analyzing +4 | -1
−2π5-\frac{2 \pi}{5} is the principal value of
View answer and explanation
Correct option: B
sin⁡−1(sin⁡7π5)\sin ^{-1}\left(\sin \frac{7 \pi}{5}\right)
Option B is correct because sin⁡−1(sin⁡x)\sin^{-1}(\sin x) has a principal value range of [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}], and sin⁡−1(sin⁡7π5)=−2π5\sin^{-1}(\sin \frac{7\pi}{5}) = -\frac{2\pi}{5}.
313 Easy
313 of 365
Application +4 | -1
If tan⁡−1(x+2)+tan⁡−1(x−2)−tan⁡−1(12)=0,  {\tan ^{ - 1}}\left( {x + 2} \right) + {\tan ^{ - 1}}\left( {x - 2} \right) - {\tan ^{ - 1}}\left( {\frac{1}{2}} \right) = 0,\; then one of the values of xx is equal to
View answer and explanation
Correct option: D
1
Option D (1) is correct because solving the equation using the tangent addition formula gives x = 1 or x = -5, and x = 1 satisfies the original equation.
314 Easy
314 of 365
Analyzing +4 | -1
If a>b>0a > b > 0, then the value of tan⁡−1(ab)+tan⁡−1(a+ba−b){\tan ^{ - 1}}\left( {\frac{a}{b}} \right) + {\tan ^{ - 1}}\left( {\frac{{a + b}}{{a - b}}} \right) depends on
View answer and explanation
Correct option: D
Neither aa nor bb
Option D (Neither aa nor bb) is correct because the expression simplifies to a constant value independent of aa and bb.
315 Easy
315 of 365
Application +4 | -1
If sin⁡−1(2x1−x2)−2sin⁡−1x=0{\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right) - 2{\sin ^{ - 1}}x = 0, then xx belongs to the interval
View answer and explanation
Correct option: B
[−1/2,  1/2]\left[ { - 1/\sqrt 2 ,\;1/\sqrt 2 } \right]
Option D ([1/2,  1]\left[ {1/\sqrt 2 ,\;1} \right]) is incorrect. Option B ([−1/2,  1/2]\left[ { - 1/\sqrt 2 ,\;1/\sqrt 2 } \right]) is correct because the equation holds true when xx is within this interval, based on the properties of inverse trigonometric functions.

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316 Easy
316 of 365
Application +4 | -1
If tan⁡−1[3sin⁡2α5+3cos⁡2α]+tan⁡−1[tan⁡α4]=λα4\tan ^{-1}\left[\frac{3 \sin 2 \alpha}{5+3 \cos 2 \alpha}\right]+\tan ^{-1}\left[\frac{\tan \alpha}{4}\right]=\frac{\lambda \alpha}{4}; where −π2<α<π2-\frac{\pi}{2}<\alpha<\frac{\pi}{2}, then find λ\lambda
View answer and explanation
Correct option: A
4
Option A (4) is correct because simplifying the given expression leads to α = λα / 4, which implies λ = 4.
317 Easy
317 of 365
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 formula for tan⁡(A−B)\tan(A-B) and simplifying, we find that A−B=30∘A - B = 30^\circ.
318 Easy
318 of 365
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, x2+1x2+2\sqrt{\frac{x^2+1}{x^2+2}}, is correct because by using trigonometric substitutions and identities, the expression can be simplified to this form.
319 Easy
319 of 365
Application +4 | -1
tan⁡[12sin⁡−1(2a1+a2)+12cos⁡−1(1−a21+a2)]  \tan \left[ {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} \right) + \frac{1}{2}{{\cos }^{ - 1}}\left( {\frac{{1 - {a^2}}}{{1 + {a^2}}}} \right)} \right]\; is equal to
View answer and explanation
Correct option: C
2a1−a2\frac{{2a}}{{1 - {a^2}}}
Option C (2a1−a2\frac{{2a}}{{1 - {a^2}}}) is correct because it simplifies to tan⁡(2tan⁡−1a)\tan(2 \tan^{-1} a) using trigonometric identities, which further simplifies to the given expression.
320 Easy
320 of 365
Application +4 | -1
If 3sin⁡−12x1+x2−4cos⁡−11+x1+x2+2tan⁡−12x1−x2=π3,3{\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} - 4{\cos ^{ - 1}}\frac{{1 + x}}{{1 + {x^2}}} + 2{\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} = \frac{\pi }{3}, then 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 it is the solution obtained by substituting x=tan⁡θx = \tan \theta and simplifying the given equation using trigonometric identities.
321 Easy
321 of 365
Application +4 | -1
The value of cos⁡−1(−12){\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) among the following, is
View answer and explanation
Correct option: B
8π3\frac{{8\pi }}{3}
Option B (8π3\frac{{8\pi }}{3}) is correct because it is one of the possible values of cos⁡−1(−12){\cos ^{ - 1}}\left( { - \frac{1}{2}} \right) obtained by considering the general solution of the cosine function.
322 Easy
322 of 365
Application +4 | -1
If sin⁡−1(6x1+9x2)=2tan⁡−1(ax)\sin ^{-1}\left(\frac{6 x}{1+9 x^2}\right)=2 \tan ^{-1}(a x), then a=a=
View answer and explanation
Correct option: A
3
Option A (3) is correct because by using the double angle formula for sine and comparing the expressions, a is found to be 3.
323 Easy
323 of 365
Application +4 | -1
cos⁡−1(1517)+2tan⁡−1(15)={\cos ^{ - 1}}\left( {\frac{{15}}{{17}}} \right) + 2{\tan ^{ - 1}}\left( {\frac{1}{5}} \right) =
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the expression simplifies to cos⁡−1(140221){\cos ^{ - 1}}(\frac{{140}}{{221}}), which is not equal to any of the other options.
324 Easy
324 of 365
Application +4 | -1
If cos⁡−135−sin⁡−145=cos⁡−1x,{\cos ^{ - 1}}\frac{3}{5} - {\sin ^{ - 1}}\frac{4}{5} = {\cos ^{ - 1}}x, then xx is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because cos⁡−135−sin⁡−145{\cos ^{ - 1}}\frac{3}{5} - {\sin ^{ - 1}}\frac{4}{5} simplifies to 0, and cos⁡−1x=0{\cos ^{ - 1}}x = 0 implies x=1x = 1.
325 Easy
325 of 365
Application +4 | -1
If −∞<x≤0- \infty < x \le 0, then cos⁡−1(1−x21+x2){\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right) equals
View answer and explanation
Correct option: B
−2tan⁡−1x- 2{\tan ^{ - 1}}x
Option B (−2tan⁡−1x- 2{\tan ^{ - 1}}x) is correct because it simplifies the given expression using trigonometric identities and the given condition on x.
326 Easy
326 of 365
Analyzing +4 | -1
The value of ′a′'a' for which ax2+sin⁡−1(x2−2x+2)+cos⁡−1(x2−2x+2)=0a{x^2} + {\sin ^{ - 1}}\left( {{x^2} - 2x + 2} \right) + {\cos ^{ - 1}}\left( {{x^2} - 2x + 2} \right) = 0 has a real solution, is
View answer and explanation
Correct option: C
−π2- \frac{\pi }{2}
Option C (−π2- \frac{\pi }{2}) is correct because substituting x=1 into the equation and solving for 'a' yields a = -π/2.
327 Easy
327 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
2π/32 \pi / 3
Option D (2π/32 \pi / 3) is correct because it is the result of evaluating the given expression using the principal values of the inverse trigonometric functions.
328 Easy
328 of 365
Application +4 | -1
If θ∈[4  π,  5  π]\theta \in \left[ {4\;\pi ,\;5\;\pi } \right], then cos⁡−1(cos⁡θ){\cos ^{ - 1}}\left( {\cos \theta } \right) equals
View answer and explanation
Correct option: A
−4  π+θ- 4\;\pi + \theta
Option C (4π−θ4\pi - \theta) is correct because, within the given range for θ\theta, the expression cos⁡−1(cos⁡θ){\cos ^{ - 1}}(\cos \theta ) simplifies to this value.
329 Easy
329 of 365
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: D
12\frac{1}{2}
Option D (12\frac{1}{2}) is correct because solving the given equation involving inverse cosines leads to q=12q = \frac{1}{2}.
330 Easy
330 of 365
Understanding +4 | -1
sin⁡−1(2x1+x2)=2tan⁡−1x\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)=2 \tan ^{-1} x for
View answer and explanation
Correct option: C
∣x∣≤1|x| \leq 1
Option C (∣x∣≤1|x| \leq 1) is correct because the identity sin⁡−1(2x1+x2)=2tan⁡−1x\sin ^{-1}(\frac{2 x}{1+x^2})=2 \tan ^{-1} x holds true only when ∣x∣≤1|x| \leq 1.
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