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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Easy

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151 Easy
151 of 365
Application +4 | -1
If cos⁡−1(x/a)+cos⁡−1(y/b)=α\cos ^{-1}(x / a)+\cos ^{-1}(y / b)=\alpha. Then x2/a2+y2/b2x^2 / a^2+y^2 / b^2 is equal to
View answer and explanation
Correct option: A
(2xy/ab)cos⁡α+sin⁡2α(2 x y / a b) \cos \alpha+\sin ^2 \alpha
The correct answer is A ((2xy/ab)cos⁡α+sin⁡2α(2 x y / a b) \cos \alpha+\sin ^2 \alpha) because using the cosine addition formula and algebraic manipulation, we can arrive at this expression.
152 Easy
152 of 365
Application +4 | -1
cos⁡−1(cos⁡(−5))+sin⁡−1(sin⁡(6))−tan⁡−1(tan⁡(12)){\cos ^{ - 1}}(\cos ( - 5)) + {\sin ^{ - 1}}(\sin (6)) - {\tan ^{ - 1}}(\tan (12)) is equal to
(The inverse trigonometric functions take the principal values)
View answer and explanation
Correct option: C
4π−11 4\pi - 11\,
Option C (4π−11 4\pi - 11\,) is correct because the inverse trigonometric functions are evaluated within their principal value ranges.
153 Easy
153 of 365
Application +4 | -1
If tan⁡−1a+tan⁡−1b=sin⁡−11−tan⁡−1c{\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\sin ^{ - 1}}1 - {\tan ^{ - 1}}c, then
View answer and explanation
Correct option: C
1a+1b+1c−1abc=0\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - \frac{1}{{abc}} = 0
Option C is correct because simplifying the given equation using the tangent addition formula and the fact that tan⁡−1a+tan⁡−1b+tan⁡−1c=π2{\tan ^{ - 1}}a + {\tan ^{ - 1}}b + {\tan ^{ - 1}}c = \frac{\pi }{2} leads to 1a+1b+1c−1abc=0\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - \frac{1}{{abc}} = 0.
154 Easy
154 of 365
Analyzing +4 | -1
The greatest and the least values of (sin⁡−1x)3+(cos⁡−1x)3{\left( {{{\sin }^{ - 1}}x} \right)^3} + {\left( {{{\cos }^{ - 1}}x} \right)^3} are respectively
View answer and explanation
Correct option: D
None of these
None of the options are correct. The greatest and least values of the expression are π332\frac{{{\pi ^3}}}{{32}} and 7π38\frac{{7{\pi ^3}}}{8} respectively, which are not listed in the options.
155 Easy
155 of 365
Application +4 | -1
cos⁡(2cos⁡−1x)+cos⁡(2sin⁡−1x)=0\cos \left(2 \cos ^{-1} x\right)+\cos \left(2 \sin ^{-1} x\right)=0 
View answer and explanation
Correct option: D
−1≤x≤1-1 \leq x \leq 1
Option B (A3A^3) is correct because A4=IA^4 = I, which implies A−1=A3A^{-1} = A^3.
156 Easy
156 of 365
Application +4 | -1
If f(x)=sin⁡−1{32x−121−x2},  −12≤x≤1f\left( x \right) = {\sin ^{ - 1}}\left\{ {\frac{{\sqrt 3 }}{2}x - \frac{1}{2}\sqrt {1 - {x^2}} } \right\},\; - \frac{1}{2} \le x \le 1, then f(x)f\left( x \right) is equal to
View answer and explanation
Correct option: B
sin⁡−1x−π6{\sin ^{ - 1}}x - \frac{\pi }{6}
Option B (sin⁡−1x−π6{\sin ^{ - 1}}x - \frac{\pi }{6}) is correct because by substituting x=sin⁡θx = \sin \theta, the expression simplifies to sin⁡−1(sin⁡(θ−π6)){\sin ^{ - 1}}(\sin(\theta - \frac{\pi}{6})), which equals θ−π6\theta - \frac{\pi}{6}.
157 Easy
157 of 365
Application +4 | -1
The value of

sin⁡(sin⁡−113+sec⁡−13)+cos⁡(tan⁡−112+tan⁡−12)\sin \left( {{{\sin }^{ - 1}}\frac{1}{3} + {{\sec }^{ - 1}}3} \right) + \cos \left( {{{\tan }^{ - 1}}\frac{1}{2} + {{\tan }^{ - 1}}2} \right) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because by using inverse trigonometric identities, the expression simplifies to sin(π/2) + cos(π/2) = 1.
158 Easy
158 of 365
Application +4 | -1
If cos⁡−1x=tan⁡−1x\cos ^{-1} x=\tan ^{-1} x, then sin⁡(cos⁡−1x)=\sin \left(\cos ^{-1} x\right)=
View answer and explanation
Correct option: B
x2x^2
Option B (x2x^2) is correct because, by setting cos⁡−1x=tan⁡−1x=θ\cos^{-1}x = \tan^{-1}x = \theta, we derive the relationship sin⁡(cos⁡−1x)=x2\sin(\cos^{-1}x) = x^2 through trigonometric identities.
159 Easy
159 of 365
Application +4 | -1
tan⁡{2tan⁡−1(15)−π4}\tan \left\{2 \tan ^{-1}\left(\frac{1}{5}\right)-\frac{\pi}{4}\right\} is equal to
View answer and explanation
Correct option: C
−717-\frac{7}{17}
Option C (−717-\frac{7}{17}) is correct because it is the simplified value of the given expression after applying the tangent addition and double angle formulas.
160 Easy
160 of 365
Understanding +4 | -1
Assertion: The function f(x)=sin⁡xf(x)=\sin x does not possess inverse if x∈Rx \in R.

Reason: The function f(x)=sin⁡xf(x)=\sin x is not one-one onto if x∈Rx \in R.
View answer and explanation
Correct option: A
Assertion is correct, reason is correct; reason is a correct explanation for assertion.
Option A is correct because the assertion is true, the reason is true, and the reason correctly explains the assertion. A function must be one-to-one and onto to have an inverse.
161 Easy
161 of 365
Application +4 | -1
The value of cos⁡−123−cos⁡−16+123\cos ^{-1} \sqrt{\frac{2}{3}}-\cos ^{-1} \frac{\sqrt{6}+1}{2 \sqrt{3}} is equal to
View answer and explanation
Correct option: C
π6\frac{\pi}{6}
Option C (π6\frac{\pi}{6}) is correct because the expression simplifies to π6\frac{\pi}{6} after applying trigonometric identities and inverse trigonometric properties.
162 Easy
162 of 365
Application +4 | -1
The value of cot⁡(∑n=150tan⁡−1(11+n+n2))\cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {\frac{1}{{1 + n + {n^2}}}} \right)} } \right) is
View answer and explanation
Correct option: A
2625\frac{{26}}{{25}}
Option A (2625\frac{{26}}{{25}}) is correct because the telescoping sum simplifies to cot⁡(tan⁡−1(50)−tan⁡−1(1))\cot \left( {{\tan }^{ - 1}}\left( {50} \right) - {{\tan }^{ - 1}}\left( 1 \right) \right), which evaluates to 2625\frac{{26}}{{25}}.
163 Easy
163 of 365
Application +4 | -1
The solution of tan⁡−1x+2cot⁡−1x=2π3{\tan ^{ - 1}}x + 2{\cot ^{ - 1}}x = \frac{{2\pi }}{3} is
View answer and explanation
Correct option: D
3\sqrt 3
Option D (3\sqrt 3) is correct because it is the solution obtained after simplifying the given trigonometric equation using inverse trigonometric identities.
164 Easy
164 of 365
Application +4 | -1
tan⁡−1(1/11)+tan⁡−1(2/12)\tan ^{-1}(1 / 11)+\tan ^{-1}(2 / 12) is equal to
View answer and explanation
Correct option: D
none of these
Option D (none of these) is correct because the sum of the two inverse tangent functions simplifies to tan⁡−1(17/65)\tan^{-1}(17/65), which is not equal to any of the other options.
165 Easy
165 of 365
Application +4 | -1
If 0<x<10 < x < 1, then

1+x2[{xcos⁡(cot⁡−1x)+sin⁡(cot⁡−1x)}2−1]1/2\sqrt {1 + {x^2}} {\left[ {{{\left\{ {x\cos \left( {{{\cot }^{ - 1}}x} \right) + \sin \left( {{{\cot }^{ - 1}}x} \right)} \right\}}^2} - 1} \right]^{1/2}} is equal to
View answer and explanation
Correct option: C
x1+x2x\sqrt {1 + {x^2}}
Option C (x1+x2x\sqrt {1 + {x^2}}) is correct because simplifying the expression using trigonometric identities leads to this result.

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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
166 Easy
166 of 365
Analyzing +4 | -1
Solution set of [sin⁡−1x]>[cos⁡−1x],  [{\sin ^{ - 1}}x] > [{\cos ^{ - 1}}x],\; white [.]denote the greatest integer function , is
View answer and explanation
Correct option: C
[sin  1,  1]\mathrm{[sin\;}1,\mathrm{\;}1]
Option C ([sin  1,  1]\mathrm{[sin\;}1,\mathrm{\;}1]) is the correct solution set because it satisfies the given inequality [sin⁡−1x]>[cos⁡−1x][{\sin ^{ - 1}}x] > [{\cos ^{ - 1}}x] considering the greatest integer function.
167 Easy
167 of 365
Application +4 | -1
If cos⁡−1x+cos⁡−1(2/3)=(π/2)\cos ^{-1} x+\cos ^{-1}(2 / 3)=(\pi / 2), the value of xx is
View answer and explanation
Correct option: B
5/3\sqrt{5} / 3
Option B (5/3\sqrt{5} / 3) is correct because by using the trigonometric identity cos⁡−1x=π2−sin⁡−1x\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x, we can find the value of x.
168 Easy
168 of 365
Application +4 | -1
The equation sin⁡−1x−cos⁡−1x=cos⁡−1(32){\sin ^{ - 1}}x - {\cos ^{ - 1}}x = {\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) has
View answer and explanation
Correct option: B
Unique solution
Option B (Unique solution) is correct because solving the given equation using trigonometric identities results in a single valid solution for x.
169 Easy
169 of 365
Understanding +4 | -1
If sec⁡−1x=cosec−1y{\sec ^{ - 1}}x = \mathrm{cose}{\mathrm{c}^{ - 1}}y, then cos⁡−11x+cos⁡−11y={\cos ^{ - 1}}\frac{1}{x} + {\cos ^{ - 1}}\frac{1}{y} =
View answer and explanation
Correct option: D
π2\frac{\pi }{2}
Option D (π2\frac{\pi }{2}) is correct because if sec⁡−1x=cosec−1y{\sec ^{ - 1}}x = \mathrm{cose}{\mathrm{c}^{ - 1}}y, then cos⁡−1(1x)=sin⁡−1(1y){\cos ^{ - 1}}(\frac{1}{x}) = {\sin ^{ - 1}}(\frac{1}{y}), and cos⁡−1(1y)+sin⁡−1(1y)=π2{\cos ^{ - 1}}(\frac{1}{y}) + {\sin ^{ - 1}}(\frac{1}{y}) = \frac{\pi }{2}.
170 Easy
170 of 365
Application +4 | -1
If cos⁡(2sin⁡−1x)=19\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}, then xx is equal to
View answer and explanation
Correct option: C
±-
Option C (±23\pm \frac{2}{3}) is correct because using the identity cos⁡(2θ)=1−2sin⁡2(θ)\cos(2\theta) = 1 - 2\sin^2(\theta) and solving for x=sin⁡(θ)x = \sin(\theta) gives ±23\pm \frac{2}{3}.
171 Easy
171 of 365
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 the expression simplifies to arctangent of 1, which equals π4\frac{\pi}{4}.
172 Easy
172 of 365
Application +4 | -1
If cos⁡−1x>sin⁡−1x,  then{\cos ^{ - 1}}x > {\sin ^{ - 1}}x,\;\mathrm{then}
View answer and explanation
Correct option: D
−1≤x<12- 1 \le x < \frac{1}{{\sqrt 2 }}
Option D (−1≤x<12- 1 \le x < \frac{1}{{\sqrt 2 }}) is correct because, given the condition cos⁡−1x>sin⁡−1x{\cos ^{ - 1}}x > {\sin ^{ - 1}}x, the solution lies in the interval −1≤x<12- 1 \le x < \frac{1}{{\sqrt 2 }}.
173 Easy
173 of 365
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 by using the property that the sum of angles in a triangle is π\pi and the formula for tan⁡−1x+tan⁡−1y\tan^{-1}x + \tan^{-1}y, we can solve for angle C.
174 Easy
174 of 365
Application +4 | -1
sin⁡(12cos⁡−145)=\sin \left( {\frac{1}{2}{{\cos }^{ - 1}}\frac{4}{5}} \right) =
View answer and explanation
Correct option: B
110\frac{1}{{\sqrt {10} }}
Option B (110\frac{1}{{\sqrt {10} }}) is correct because using the half-angle formula for sine, the expression simplifies to 1−cos⁡θ2=110\sqrt{\frac{1 - \cos \theta}{2}} = \frac{1}{\sqrt{10}}.
175 Easy
175 of 365
Application +4 | -1
If e[sin⁡2α+sin⁡4α+sin⁡6α+...∞]log⁡e2{e^{[{{\sin }^2}\alpha + {{\sin }^4}\alpha + {{\sin }^6}\alpha + ...\infty ]{{\log }_e}2}} is a root of equation x2−9x+8=0,{x^2} - 9x + 8 = 0,where 0<α<π2,\alpha < \frac{\pi }{2}, then the principle value of sin⁡−1sin⁡(2π3){\sin ^{ - 1}}\sin \left( {\frac{{2\pi }}{3}} \right) is
View answer and explanation
Correct option: A
α\alpha
Option A (α\alpha) is correct because, after simplifying the given expression and solving for α\alpha, the principal value of sin⁡−1sin⁡(2π3){\sin ^{ - 1}}\sin \left( {\frac{{2\pi }}{3}} \right) is equal to α\alpha.
176 Easy
176 of 365
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 is correct because solving the equation sin⁡−1x−cos⁡−1x=π6{\sin ^{ - 1}}x - {\cos ^{ - 1}}x = \frac{\pi }{6} using trigonometric identities leads to x=32x = \frac{{\sqrt 3 }}{2}.
177 Easy
177 of 365
Application +4 | -1
The value of cos⁡(2cos⁡−10.8)\cos \left( {2{{\cos }^{ - 1}}0.8} \right) is
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because the value of the expression is 0.28, which is not among the given options.
178 Easy
178 of 365
Analyzing +4 | -1
The equation 2cos⁡−1x+sin⁡−1x=11π62{\cos ^{ - 1}}x + {\sin ^{ - 1}}x = \frac{{11\pi }}{6} has
View answer and explanation
Correct option: A
No solution
Option A (tan⁡−1(n2+nn2+n+2){\tan ^{ - 1}}\left( {\frac{{{n^2} + n}}{{{n^2} + n + 2}}} \right)) is correct because the summation telescopes, resulting in the given expression.
179 Easy
179 of 365
Application +4 | -1
If 12≤x≤1\frac{1}{2} \le x \le 1, then sin⁡−1(3x−4x3){\sin ^{ - 1}}\left( {3x - 4{x^3}} \right) equals
View answer and explanation
Correct option: B
π−3sin⁡−1x\pi - 3{\sin ^{ - 1}}x
Option B (π−3sin⁡−1x\pi - 3{\sin ^{ - 1}}x) is correct because within the given range of x, the expression sin⁡−1(3x−4x3){\sin ^{ - 1}}(3x - 4{x^3}) simplifies to π−3sin⁡−1x\pi - 3{\sin ^{ - 1}}x.
180 Easy
180 of 365
Application +4 | -1
If sin⁡−1x+sin⁡−1y+sin⁡−1z=3π2{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \frac{{3\pi }}{2}, then the value of x100+y100+z100−9x101+y101+z101{x^{100}} + {y^{100}} + {z^{100}} - \frac{9}{{{x^{101}} + {y^{101}} + {z^{101}}}}, is
View answer and explanation
Correct option: A
0
Option A is correct because when sin⁡−1x+sin⁡−1y+sin⁡−1z=3π2{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \frac{{3\pi }}{2}, it implies x=y=z=1x=y=z=1, and therefore x100+y100+z100−9x101+y101+z101=0{x^{100}} + {y^{100}} + {z^{100}} - \frac{9}{{{x^{101}} + {y^{101}} + {z^{101}}}} = 0.
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