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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Mixed difficulty

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241 Easy
241 of 605
Application +4 | -1
tan⁡(2tan⁡−115+sec⁡−152+2tan⁡−118)\tan \left(2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{\sqrt{5}}{2}+2 \tan ^{-1} \frac{1}{8}\right) is equal to :
View answer and explanation
Correct option: B
2
Option B (2) is correct because by applying trigonometric identities and simplification, the expression reduces to tan(tan-1(2)) which equals 2.
242 Easy
242 of 605
Application +4 | -1
The solution of tan⁡−12x+tan⁡−13x=π4{\tan ^{ - 1}}2x + {\tan ^{ - 1}}3x = \frac{\pi }{4} is
View answer and explanation
Correct option: A
16\frac{1}{6}
Option A (16\frac{1}{6}) is correct because solving the equation tan⁡−12x+tan⁡−13x=π4{\tan ^{ - 1}}2x + {\tan ^{ - 1}}3x = \frac{\pi }{4} yields x=16x = \frac{1}{6} as the only valid solution.
243 Easy
243 of 605
Application +4 | -1
The value of cos⁡−1(cos⁡(−17π/5))\cos ^{-1}(\cos (-17 \pi / 5)), is equal to
View answer and explanation
Correct option: B
3π/53 \pi / 5
Option B (3π/53 \pi / 5) is correct because cos⁡−1(cos⁡(−17π/5))\cos^{-1}(\cos(-17\pi/5)) simplifies to cos⁡−1(cos⁡(3π/5))\cos^{-1}(\cos(3\pi/5)) which equals 3π/53\pi/5 since 3π/53\pi/5 is within the range of cos⁡−1\cos^{-1}.
244 Medium
244 of 605
Application +4 | -1
If sin⁡−1x≤cos⁡−1x\sin ^{-1} x \leq \cos ^{-1} x, then x∈……x \in \ldots \ldots
View answer and explanation
Correct option: B
[−1,12]\left[-1, \frac{1}{\sqrt{2}}\right]
Option B ([−1,12]\left[-1, \frac{{1}}{{\sqrt{{2}}}}\right]) is correct because solving the inequality sin⁡−1x≤cos⁡−1x\sin^{-1}x \le \cos^{-1}x leads to this interval.
245 Easy
245 of 605
Application +4 | -1
If 2sin⁡−1x=sin⁡−1(2x1−x2),{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right), then xx is equal to
View answer and explanation
Correct option: C
[−12,12]\left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]
Option C ([−12,12]\left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]) is correct because it represents the range of x for which the given equation holds true, considering the domain of the inverse sine function.
246 Easy
246 of 605
Application +4 | -1
The principal value of cos⁡−1(cos⁡2π3)+sin⁡−1(sin⁡2π3)\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right) is
View answer and explanation
Correct option: A
π\pi
Option A (π\pi) is correct because it is the result of simplifying the expression using the properties of inverse trigonometric functions.
247 Easy
247 of 605
Application +4 | -1
The value of cot⁡−1xy+1x−y+cot⁡−1yz+1y−z+cot⁡−1zx+1z−x{\cot ^{ - 1}}\frac{{xy + 1}}{{x - y}} + {\cot ^{ - 1}}\frac{{yz + 1}}{{y - z}} + {\cot ^{ - 1}}\frac{{zx + 1}}{{z - x}} is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the expression simplifies to zero due to the telescoping nature of the inverse cotangent terms.
248 Easy
248 of 605
Application +4 | -1
If tan⁡−1(13)+tan⁡−1(34)−tan⁡−1(x3)=0{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{3}{4}} \right) - {\tan ^{ - 1}}\left( {\frac{x}{3}} \right) = 0, then xx is equal to
View answer and explanation
Correct option: D
133\frac{{13}}{3}
Option D (133\frac{{13}}{3}) is correct because by applying the 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), we can solve for x.
249 Medium
249 of 605
Application +4 | -1
cos⁡[tan⁡−113+tan⁡−112]\cos \left[\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{2}\right] is equal to
View answer and explanation
Correct option: A
12\frac{1}{\sqrt{2}}
Option A (12\frac{1}{\sqrt{2}}) is correct because by using the formula for tan⁡(A+B)\tan(A+B) and then finding the cosine, we arrive at the correct value.
250 Easy
250 of 605
Application +4 | -1
tan⁡−134+tan⁡−135−cot⁡−18\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{3}{5}-\cot ^{-1} 8 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 tangent addition and subtraction formulas results in tan⁡−1(1)\tan^{-1}(1), which equals π4\frac{{\pi}}{{4}}.
251 Hard
251 of 605
Analyzing +4 | -1
The value of ∑r=0∞tan⁡−1(11+r+r2)  \sum \limits_{r = 0}^\infty {\tan ^{ - 1}}\left( {\frac{1}{{1 + r + {r^2}}}} \right)\; is equal to
View answer and explanation
Correct option: A
π2\frac{\pi }{2}
Option A (π2\frac{\pi }{2}) is correct because the sum telescopes to tan⁡−1(∞)\tan^{-1}(\infty), which equals π2\frac{\pi }{2}.
252 Easy
252 of 605
Understanding +4 | -1
sec⁡−1[sec⁡(−30∘)]\sec ^{-1}\left[\sec \left(-30^{\circ}\right)\right] is equal to
View answer and explanation
Correct option: C
30∘30^{\circ}
Option C (30∘30^{\circ}) is correct because sec⁡(−30∘)=sec⁡(30∘)\sec(-30^{\circ}) = \sec(30^{\circ}) and the range of sec⁡−1x\sec^{-1}x is [0,π][0, \pi], excluding π/2\pi/2.
253 Easy
253 of 605
Application +4 | -1
If cos⁡−1x+cos⁡−1y+cos⁡−1z+cos⁡−1t=4π{\cos ^{ - 1}}x + {\cos ^{ - 1}}y + {\cos ^{ - 1}}z + {\cos ^{ - 1}}t = 4\pi, then the value of x2+y2+z2+t2{x^2} + {y^2} + {z^2} + {t^2} is
View answer and explanation
Correct option: C
4
Option C (4) is correct because the given condition cos⁡−1x+cos⁡−1y+cos⁡−1z+cos⁡−1t=4π{\cos ^{ - 1}}x + {\cos ^{ - 1}}y + {\cos ^{ - 1}}z + {\cos ^{ - 1}}t = 4\pi implies x=y=z=t=−1x = y = z = t = -1, and thus x2+y2+z2+t2=4{x^2} + {y^2} + {z^2} + {t^2} = 4.
254 Easy
254 of 605
Application +4 | -1
If −1<x<1- 1 < x < 1, then tan⁡−1(2x1−x2){\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right) equals
View answer and explanation
Correct option: A
2tan⁡−1x2{\tan ^{ - 1}}x
Option A (2tan⁡−1x2{\tan ^{ - 1}}x) is correct because using the identity tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta = \frac{{2\tan \theta}}{{1 - {{\tan }^2}\theta}} and the given condition −1<x<1-1 < x < 1, we can simplify the expression.
255 Easy
255 of 605
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.

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256 Easy
256 of 605
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.
257 Medium
257 of 605
Application +4 | -1
The value of cos⁡−1(−12)−2sin⁡−1(12)+3cos⁡−1(−12)−4tan⁡−1(−1)\cos ^{-1}\left(-\frac{1}{2}\right)-2 \sin ^{-1}\left(\frac{1}{2}\right)+3 \cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)-4 \tan ^{-1}(-1) is equal to
View answer and explanation
Correct option: D
25π/1225 \pi / 12
Option D (25π12\frac{{25 \pi}}{{12}}) is correct because it correctly evaluates each inverse trigonometric function and performs the arithmetic.
258 Medium
258 of 605
Application +4 | -1
Considering the principal values of inverse trigonometric functions, the positive real values of 'xx' satisfying tan⁡−1x+tan⁡−1(2x)=π4\tan ^{-1} x+\tan ^{-1}(2 x)=\frac{\pi}{4} is
View answer and explanation
Correct option: C
17−34\frac{\sqrt{17}-3}{4}
Option C (17−34\frac{\sqrt{17}-3}{4}) is correct because solving the equation tan⁡−1x+tan⁡−1(2x)=π4\tan^{-1} x + \tan^{-1} (2x) = \frac{\pi}{4} leads to a quadratic equation, and the positive solution is 17−34\frac{\sqrt{17}-3}{4}.
259 Medium
259 of 605
Application +4 | -1
The value of tan⁡2(sec⁡−12)+cot⁡2(cosec⁡−13)\tan ^2\left(\sec ^{-1} 2\right)+\cot ^2\left(\operatorname{cosec}^{-1} 3\right) is
View answer and explanation
Correct option: D
11
Option D (11) is correct because by converting sec⁡−12\sec^{-1}2 to tan⁡−1(3)\tan^{-1}(\sqrt{3}) and csc⁡−13\csc^{-1}3 to cot⁡−1(22)\cot^{-1}(2\sqrt{2}), and then squaring, the result is 11.
260 Easy
260 of 605
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.
261 Easy
261 of 605
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.
262 Easy
262 of 605
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&nbsp;
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.
263 Easy
263 of 605
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}.
264 Easy
264 of 605
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.
265 Easy
265 of 605
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.
266 Medium
266 of 605
Application +4 | -1
tan⁡[π4+12cos⁡−1ab]+tan⁡[π4−12cos⁡−1ab]\tan \left[\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} \frac{a}{b}\right]+\tan \left[\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} \frac{a}{b}\right] equal to
View answer and explanation
Correct option: B
2ba\frac{2 b}{a}
Option B (2ba\frac{2 b}{a}) is correct because simplifying the given expression using trigonometric identities leads to 2ba\frac{2b}{a}.
267 Medium
267 of 605
Application +4 | -1
The value of tan⁡−1ab−tan⁡−1a−ba+b\tan ^{-1} \frac{a}{b}-\tan ^{-1} \frac{a-b}{a+b} for ∣a∣<∣b∣|\mathrm{a}|<|\mathrm{b}| is
View answer and explanation
Correct option: B
π4\frac{\pi}{4}
Option B (π4\frac{\pi}{4}) is correct because the expression simplifies to tan⁡−1(1)\tan^{-1}(1), which equals π4\frac{\pi}{4}, using the inverse tangent subtraction formula.
268 Easy
268 of 605
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.
269 Easy
269 of 605
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.
270 Medium
270 of 605
Analyzing +4 | -1
Let (a,b)⊂(0,2π)(a, b) \subset(0,2 \pi) be the largest interval for which sin⁡−1(sin⁡θ)−cos⁡−1(sin⁡θ)>0,θ∈(0,2π)\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>0, \theta \in(0,2 \pi), holds. If αx2+βx+sin⁡−1(x2−6x+10)+cos⁡−1(x2−6x+10)\alpha x^{2}+\beta x+\sin ^{-1}\left(x^{2}-6 x+10\right)+\cos ^{-1}\left(x^{2}-6 x+10\right) =0=0 and α−β=b−a\alpha-\beta=b-a, then α\alpha is equal to :
View answer and explanation
Correct option: D
π12\frac{\pi}{12}
Option D (π12\frac{\pi}{12}) is correct because by solving the given conditions, we find that α=π12\alpha = \frac{\pi}{12}.
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