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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Easy

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331 Easy
331 of 365
Application +4 | -1
If cos⁡−1x+cos⁡−1y=π2  {\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \frac{\pi }{2}\; and tan⁡−1x−tan⁡−1y=0,  {\tan ^{ - 1}}x - {\tan ^{ - 1}}y = 0,\; then x2+xy+y2  {x^2} + xy + {y^2}\; is equal to
View answer and explanation
Correct option: C
32\frac{3}{2}
Option C is correct because given the conditions cos⁡−1x+cos⁡−1y=π2{\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \frac{\pi }{2} and tan⁡−1x−tan⁡−1y=0{\tan ^{ - 1}}x - {\tan ^{ - 1}}y = 0, it follows that x2+xy+y2=32{x^2} + xy + {y^2} = \frac{3}{2}.
332 Easy
332 of 365
Application +4 | -1
If tan⁡−1k−tan⁡−13=tan⁡−1113\tan ^{-1} k-\tan ^{-1} 3=\tan ^{-1} \frac{1}{13}, then k=k=
View answer and explanation
Correct option: A
4
Option A (4) is correct because by applying the tangent subtraction formula and solving for k, the value of k is found to be 4.
333 Easy
333 of 365
Application +4 | -1
If cos⁡−1xa+cos⁡−1yb=5π12\cos ^{-1} \frac{x}{a}+\cos ^{-1} \frac{y}{b}=\frac{5 \pi}{12} and sin⁡−1xa−sin⁡−1yb=π12\sin ^{-1} \frac{x}{a}-\sin ^{-1} \frac{y}{b}=\frac{\pi}{12} then x2a2+y2b2\frac{x^2}{a^2}+\frac{y^2}{b^2} is
View answer and explanation
Correct option: D
54\frac{5}{4}
Option B (32e\frac{3}{2} \sqrt{e}) is correct because by solving the differential equation and applying the given condition, we can find y(4).
334 Easy
334 of 365
Application +4 | -1
If 12sin⁡−1[3sin⁡2θ5+4cos⁡2θ]=tan⁡−1x\frac{1}{2} \sin ^{-1}\left[\frac{3 \sin 2 \theta}{5+4 \cos 2 \theta}\right]=\tan ^{-1} x, then x=x=
View answer and explanation
Correct option: C
(1/3)tan⁡θ(1 / 3) \tan \theta
Option C ((1/3)tan(theta)) is correct because, by simplifying the expression inside the inverse sine function and using trigonometric substitutions, we find that x = (1/3)tan(theta).
335 Easy
335 of 365
Application +4 | -1
If 2tan⁡−1(cos⁡x)=tan⁡−1(2  cosec  x)2{\tan ^{ - 1}}\left( {\cos x} \right) = {\tan ^{ - 1}}\left( {2\;\mathrm{cosec}\;x} \right), then the value of xx is
View answer and explanation
Correct option: B
π4\frac{\pi }{4}
Option B (π4\frac{\pi}{{4}}) is correct because solving the given trigonometric equation involving inverse tangent and cosecant functions leads to x = π4\frac{\pi}{{4}}.
336 Easy
336 of 365
Application +4 | -1
If x∈[−1,0)x \in[-1,0) then cos⁡−1(2x2−1)−2sin⁡−1x\cos ^{-1}\left(2 x^2-1\right)-2 \sin ^{-1} x equals
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the expression simplifies to a constant value, but that value is not among the given options.
337 Easy
337 of 365
Application +4 | -1
tan⁡−1mn−tan⁡−1m−nm+n{\tan ^{ - 1}}\frac{m}{n} - {\tan ^{ - 1}}\frac{{m - n}}{{m + n}} is equal to
View answer and explanation
Correct option: A
π4\frac{\pi }{4}
Option A (π4\frac{\pi}{4}) is correct because simplifying the expression using the tangent subtraction formula leads to π4\frac{\pi}{4}.
338 Easy
338 of 365
Application +4 | -1
sin⁡(cos⁡−1(45))=\sin \left(\cos ^{-1}\left(\frac{4}{5}\right)\right)=
View answer and explanation
Correct option: C
35\frac{3}{5}
Option C (3/5) is correct because sin(cos-1(4/5)) can be simplified by considering a right triangle where cos(θ) = 4/5, which implies sin(θ) = 3/5.
339 Easy
339 of 365
Application +4 | -1
cot⁡{cos⁡−1(725)}  =\cot \left\{ {{{\cos }^{ - 1}}\left( {\frac{7}{{25}}} \right)} \right\}\; =
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because the value of the expression simplifies to 7/24, which is not among the given options.
340 Easy
340 of 365
Application +4 | -1
The value of cos⁡(2cos⁡−1x+sin⁡−1x)  at  x=15  \cos (2{\cos ^{ - 1}}x + {\sin ^{ - 1}}x)\;\mathrm{at}\;x = \frac{1}{5}\; is
View answer and explanation
Correct option: D
−265- \frac{{2\sqrt 6 }}{5}
Option D (−265- \frac{{2\sqrt 6 }}{5}) is correct because it involves simplifying the expression using trigonometric identities and substituting the given value of x.
341 Easy
341 of 365
Application +4 | -1
The value of tan⁡−112+tan⁡−113+tan⁡−178{\tan ^{ - 1}}\frac{1}{2} + {\tan ^{ - 1}}\frac{1}{3} + {\tan ^{ - 1}}\frac{7}{8} is
View answer and explanation
Correct option: C
tan⁡−115{\tan ^{ - 1}}15
The correct answer is C (tan⁡−115{\tan ^{ - 1}}15) because it results from applying the arctangent addition formula twice to simplify the expression.
342 Easy
342 of 365
Application +4 | -1
A continuous even periodic function ff with period 8 is such that f(0)=0,f(1)=−2,f(2)=1,f(3)=2,f(4)=3f(0)=0, f(1)=-2, f(2)=1, f(3)=2, f(4)=3 then the value of tan⁡−1(tan⁡(f(−5)+f(20)+cos⁡−1(f(−10)+f(17))))\tan ^{-1}\left(\tan \left(f(-5)+f(20)+\cos ^{-1}(f(-10)+f(17))\right)\right) is equal to:
View answer and explanation
Correct option: B
5−2π5-2 \pi
Option B (5−2π5-2 \pi) is correct because after simplification using the properties of trigonometric functions and inverse trigonometric functions, the expression evaluates to 5−2π5-2 \pi.
343 Easy
343 of 365
Application +4 | -1
The value of cos⁡−1(−sin⁡7π6)\cos ^{-1}\left(-\sin \frac{7 \pi}{6}\right)
View answer and explanation
Correct option: C
π3\frac{\pi}{3}
Option C (π3\frac{{\pi}}{{3}}) is correct because it correctly evaluates the sine function and then applies the inverse cosine function.
344 Easy
344 of 365
Analyzing +4 | -1
If sin⁡−1x+sin⁡−1y+sin⁡−1z=π,  then  x4+y4+z4+4x2y2z2=k(x2y2+y2z2+z2x2){\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \pi ,\;\mathrm{then}\;{x^4} + {y^4} + {z^4} + 4{x^2}{y^2}{z^2} = k\left( {{x^2}{y^2} + {y^2}{z^2} + {z^2}{x^2}} \right) Where kk is equal to
View answer and explanation
Correct option: B
2
Option B (2) is correct because by manipulating the given equation and using trigonometric identities, we can show that the constant k in the expression is equal to 2.
345 Easy
345 of 365
Application +4 | -1
If tan⁡−1(ax)+tan⁡−1(bx)=π2{\tan ^{ - 1}}\left( {\frac{a}{x}} \right) + {\tan ^{ - 1}}\left( {\frac{b}{x}} \right) = \frac{\pi }{2}, then xx is equal to
View answer and explanation
Correct option: A
ab\sqrt {ab}
Option A (ab\sqrt {ab}) is correct because it is the solution obtained by applying the tangent addition formula and solving the resulting equation.

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346 Easy
346 of 365
Application +4 | -1
The value of cos⁡[2tan⁡−1(−7)]\cos \left[ {2{{\tan }^{ - 1}}\left( { - 7} \right)} \right] is
View answer and explanation
Correct option: D
−2425- \frac{{24}}{{25}}
Option D (−2425- \frac{{24}}{{25}}) is correct because it is the result of applying the double angle formula for cosine in terms of tangent and simplifying the expression.
347 Easy
347 of 365
Analyzing +4 | -1
If cos⁡−1x+cos⁡−1y=π2  {\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \frac{\pi }{2}\; and tan⁡−1x−tan⁡−1y=0,  {\tan ^{ - 1}}x - {\tan ^{ - 1}}y = 0,\; then x2+xy+y2  {x^2} + xy + {y^2}\; is equal to
View answer and explanation
Correct option: C
32\frac{3}{2}
Option C (3/2) is correct because from the given equations, we can deduce that x = y = 1/sqrt(2), and substituting these values into the expression x² + xy + y² yields 3/2.
348 Easy
348 of 365
Application +4 | -1
If sin⁡(sin⁡−115+cos⁡−1x)=1\sin \left( {{{\sin }^{ - 1}}\frac{1}{5} + {{\cos }^{ - 1}}x} \right) = 1, then xx is equal to
View answer and explanation
Correct option: D
1/51/5
Option D is correct because solving the equation sin⁡(sin⁡−115+cos⁡−1x)=1\sin \left( {{{\sin }^{ - 1}}\frac{1}{5} + {{\cos }^{ - 1}}x} \right) = 1 using trigonometric identities leads to x=15x = \frac{1}{5}.
349 Easy
349 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}
Option B (±53\pm \frac{{\sqrt 5 }}{3}) is correct because by setting cot⁡−112=ϕ{\cot ^{ - 1}}\frac{1}{2} = \phi and cos⁡−1x=θ{\cos ^{ - 1}}x = \mathrm{\theta }, we can solve for x using trigonometric identities.
350 Easy
350 of 365
Application +4 | -1
cosec[2cot⁡−1(5)+cos⁡−1(45)]\mathrm{cosec}\left[ {2{{\cot }^{ - 1}}(5) + {{\cos }^{ - 1}}\left( {\frac{4}{5}} \right)} \right] is equal to
View answer and explanation
Correct option: C
6556 \frac{{65}}{{56}}\,
The correct answer is C (6556\frac{{65}}{{56}}) because it correctly simplifies the trigonometric expression using inverse trigonometric identities.
351 Easy
351 of 365
Analyzing +4 | -1
If ∑i=120sin⁡−1xi=10π,\sum \limits_{i = 1}^{20} {\sin ^{ - 1}}{x_i} = 10\pi , then ∑i=120xi\sum \limits_{i = 1}^{20} {x_i} is equal to
View answer and explanation
Correct option: A
20
Option A (20) is correct because the maximum value of sin⁡−1x{\sin ^{ - 1}}x is π2\frac{\pi}{2}. Therefore, each term in the summation must be π2\frac{\pi}{2}, implying each xix_i is 1.
352 Easy
352 of 365
Application +4 | -1
If tan⁡−1(x−1)+tan⁡−1x+tan⁡−1(x+1)=tan⁡−13x{\tan ^{ - 1}}\left( {x - 1} \right) + {\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( {x + 1} \right) = {\tan ^{ - 1}}3x, then xx is
View answer and explanation
Correct option: D
0,  ±120,\; \pm \frac{1}{2}
Option D (0,±120, \pm \frac{1}{2}) is correct because these are the solutions obtained by simplifying the given equation using the arctangent addition formula.
353 Easy
353 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: tan⁡−1x−tan⁡−1y=tan⁡−1(x−y1+xy){\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right).
354 Easy
354 of 365
Application +4 | -1
If sin⁡(sin⁡−115+cos⁡−1x)=1\sin \left( {{{\sin }^{ - 1}}\frac{1}{5} + {{\cos }^{ - 1}}x} \right) = 1, then the value of xx is
View answer and explanation
Correct option: D
1/51/5
Option D (1/5) is correct because it is derived from using the identity sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} and solving for x.
355 Easy
355 of 365
Analyzing +4 | -1
If sin⁡−1(2x1+x2)+cos⁡−1(1−x21+x2)=4tan⁡−1x{\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right) + {\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right) = 4{\tan ^{ - 1}}x, then
View answer and explanation
Correct option: C
x∈[0,  1]x \in \left[ {0,\;1} \right]
Option C (x∈[0,  1]x \in \left[ {0,\;1} \right]) is correct because it is the interval where both inverse trigonometric functions are defined and the equation holds true.
356 Easy
356 of 365
Analyzing +4 | -1
Which one of the following is correct?
View answer and explanation
Correct option: A
tan⁡1>tan⁡−11\tan 1 > {\tan ^{ - 1}}1
Option A (tan⁡1>tan⁡−11\tan 1 > {\tan ^{ - 1}}1) is correct because 1 radian is greater than 45 degrees, so tan⁡1>1\tan 1 > 1, and tan⁡−11=π4<1{\tan ^{ - 1}}1 = \frac{\pi }{4} < 1.
357 Easy
357 of 365
Application +4 | -1
A value of xx for which sin⁡(cot⁡−1(1+x))=cos⁡(tan⁡−1x)\sin \left(\cot ^{-1}(1+x)\right)=\cos \left(\tan ^{-1} x\right), is
View answer and explanation
Correct option: A
−12-\frac{1}{2}
Option A (−12-\frac{{1}}{{2}}) is correct because substituting x=−12x = -\frac{{1}}{{2}} into the given equation satisfies the equation after simplification.
358 Easy
358 of 365
Analyzing +4 | -1
If the mapping f(x)=ax+b,a>0f\left( x \right) = ax + b,a > 0 maps [−1,1- 1,1] onto [0, 2] then cot⁡[cot⁡−17+cot⁡−18+cot⁡−118]\cot [{\cot ^{ - 1}}7 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18] is equal to
View answer and explanation
Correct option: D
f(2)f\left( 2 \right)
Option D (f(2)f\left( 2 \right)) is correct because after finding the function f(x) and simplifying the cotangent expression, the result equals 3, which is f(2).
359 Easy
359 of 365
Application +4 | -1
tan⁡−13−cot⁡−1(−3)\tan ^{-1} \sqrt{3}-\cot ^{-1}(-\sqrt{3}) is equal to
View answer and explanation
Correct option: B
−π2-\frac{\pi}{2}
Option B (−π2-\frac{{\pi}}{{2}}) is correct because it is the result of subtracting the principal values of the inverse cotangent from the inverse tangent.
360 Easy
360 of 365
Application +4 | -1
sin⁡[3sin⁡−1(15)]\sin \left[ {3{{\sin }^{ - 1}}\left( {\frac{1}{5}} \right)} \right] is equal to
View answer and explanation
Correct option: A
71125\frac{{71}}{{125}}
Option A (71/125) is correct because it is the result of applying the triple angle identity for sine and simplifying the expression.
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