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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Mixed difficulty

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91 Easy
91 of 605
Application +4 | -1
Two angles of a triangle are cot⁡−12{\cot ^{ - 1}}2 and cot⁡−13{\cot ^{ - 1}}3. Then , the third angle is
View answer and explanation
Correct option: B
3  π4\frac{{3\;\pi }}{4}
Option B (3  π4\frac{{3\;\pi }}{4}) is correct because the sum of the given angles is π4\frac{\pi}{4}, and the third angle is π−π4=3π4\pi - \frac{\pi}{4} = \frac{3\pi}{4}.
92 Medium
92 of 605
Application +4 | -1
cot⁡−1xy+1x−y+cot⁡−1yz+1y−z+cot⁡−1zx+1z−x;x>y>z>0\cot ^{-1} \frac{x y+1}{x-y}+\cot ^{-1} \frac{y z+1}{y-z}+\cot ^{-1} \frac{z x+1}{z-x} ; x > y > z > 0 and x,y,zx, y, z distinct is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because the expression simplifies to 0 using the inverse tangent subtraction formula and the given conditions.
93 Medium
93 of 605
Application +4 | -1
2tan⁡−1(cos⁡x)=tan⁡−1(cosec⁡2x)2 \tan ^{-1}(\cos x)=\tan ^{-1}\left(\operatorname{cosec}^2 x\right), then x=x=
View answer and explanation
Correct option: D
π3\frac{\pi}{3}
Option D (π3\frac{\pi}{3}) is correct because solving the trigonometric equation 2tan⁡−1(cos⁡x)=tan⁡−1(cosec⁡2x)2 \tan^{-1}(\cos x) = \tan^{-1}(\operatorname{cosec}^2 x) leads to x=π3x = \frac{\pi}{3}.
94 Medium
94 of 605
Analyzing +4 | -1
Let SS be solution set for values of xx satisfying cos⁡−1(2x)+cos⁡−11−x2=π\cos ^{-1}(2 x)+\cos ^{-1} \sqrt{1-x^2}=\pi, then ∑x∈S2sin⁡−1(x2−1)\sum\limits_{{x \in S}} 2 \sin ^{-1}\left(x^2-1\right) is equal to:
View answer and explanation
Correct option: B
−sin⁡−1(2425)-\sin ^{-1}\left(\frac{24}{25}\right)
Option B (−sin⁡−1(2425)-\sin ^{-1}\left(\frac{24}{25}\right)) is correct because solving the equation for xx and substituting it back into the expression yields −sin⁡−1(2425)-\sin ^{-1}\left(\frac{24}{25}\right).
95 Medium
95 of 605
Application +4 | -1
If cos⁡−1x−sin⁡−1x=0\cos ^{-1} x-\sin ^{-1} x=0, then xx is equal to
View answer and explanation
Correct option: C
12\frac{1}{\sqrt{2}}
Option C (12\frac{{1}}{{\sqrt{{2}}}}) is correct because it is the value of x that satisfies the equation cos⁡−1x−sin⁡−1x=0\cos^{-1} x - \sin^{-1} x = 0.
96 Medium
96 of 605
Application +4 | -1
cos⁡[2cos⁡−115+sin⁡−115]\cos \left[2 \cos ^{-1} \frac{1}{5}+\sin ^{-1} \frac{1}{5}\right] is equal to
View answer and explanation
Correct option: B
−265-\frac{2 \sqrt{6}}{5}
Option B (−265-\frac{2 \sqrt{6}}{5}) is correct because using the identity sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} and then simplifying the cosine expression leads to the correct answer.
97 Medium
97 of 605
Application +4 | -1
The smallest and the largest values of tan⁡−1(1−x1+x),  0≤x≤1{\tan ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right),\;0 \le x \le 1 are
View answer and explanation
Correct option: B
0,π40,\frac{\pi }{4}
Option B (0,π40,\frac{\pi }{4}) is correct because it represents the range of the given function when xx varies from 0 to 1.
98 Medium
98 of 605
Understanding +4 | -1
cos⁡−1a−xa−b=sin⁡−1x−ba−b\cos ^{-1} \sqrt{\frac{a-x}{a-b}}=\sin ^{-1} \sqrt{\frac{x-b}{a-b}} is possible if
View answer and explanation
Correct option: A
a>x>ba > x > b
Option A (a>x>ba > x > b) is the correct condition for the given trigonometric equation to be possible, ensuring the arguments of the inverse trigonometric functions are within the valid range.
99 Easy
99 of 605
Application +4 | -1
If cot⁡(cos⁡−1x)=sec⁡(tan⁡−1ab2−a2)\cot \left( {{{\cos }^{ - 1}}x} \right) = \sec \left( {{{\tan }^{ - 1}}\frac{a}{{\sqrt {{b^2} - {a^2}} }}} \right), then xx is equal to
View answer and explanation
Correct option: A
b2b2−a2\frac{b}{{\sqrt {2{b^2} - {a^2}} }}
Option A (b2b2−a2\frac{b}{{\sqrt {2{b^2} - {a^2}} }}) is correct because by converting both sides to cotangent and secant expressions, we can solve for x.
100 Easy
100 of 605
Application +4 | -1
If sin⁡−1(3x)+sin⁡−1(4x)=π2{\sin ^{ - 1}}\left( {\frac{3}{x}} \right) + {\sin ^{ - 1}}\left( {\frac{4}{x}} \right) = \frac{\pi }{2}, then xx is equal to
View answer and explanation
Correct option: B
5
Option B (5) is correct because solving the equation using trigonometric identities and squaring both sides leads to x = 5.
101 Hard
101 of 605
Analyzing +4 | -1
If θ  \mathrm{\theta }\; and ϕ\phi are the roots of the equation 8x2+22x+5=0,8{x^2} + 22x + 5 = 0, then
View answer and explanation
Correct option: C
Both tan⁡−1θ{\tan ^{ - 1}}\mathrm{\theta } and tan⁡−1ϕ{\tan ^{ - 1}}\phi are equal
Option C (Both tan⁡−1θ{\tan ^{ - 1}}\theta and tan⁡−1ϕ{\tan ^{ - 1}}\phi are equal) is correct because both roots of the quadratic equation are real numbers, and the arctangent function is defined for all real numbers.
102 Easy
102 of 605
Application +4 | -1
cos⁡−1(12)+2sin⁡−1(12){\cos ^{ - 1}}\left( {\frac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\frac{1}{2}} \right) is equal to
View answer and explanation
Correct option: C
2π3\frac{{2\pi }}{3}
Option C (2π3\frac{{2\pi }}{{3}}) is correct because it is the result of directly substituting the known values of the inverse trigonometric functions and simplifying.
103 Medium
103 of 605
Application +4 | -1
If A=tan⁡−117,B=tan⁡−113A=\tan ^{-1} \frac{1}{7}, B=\tan ^{-1} \frac{1}{3}, then
View answer and explanation
Correct option: D
cos⁡2A=sin⁡4B\cos 2 A=\sin 4 B
Option D (cos⁡2A=sin⁡4B\cos 2 A=\sin 4 B) is correct because, given the values of A and B, it can be shown that cos⁡2A\cos 2A and sin⁡4B\sin 4B both equal 2425\frac{{24}}{{25}}.
104 Easy
104 of 605
Application +4 | -1
If 2 tan⁡−1(cos  x)=tan⁡−1(2  cosec  x),  {\tan ^{ - 1}}\left( {\mathrm{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 tangents leads to x=π4x = \frac{\pi}{4}.
105 Easy
105 of 605
Application +4 | -1
If tan⁡(cos⁡−1x)=sin⁡cot⁡−1(1/2)\tan \left(\cos ^{-1} x\right)=\sin \cot ^{-1}(1 / 2), then the value of xx is
View answer and explanation
Correct option: C
5/3\sqrt{5} / 3
Option C (5/3\sqrt{5}/3) is correct because it is the solution obtained by converting the inverse trigonometric functions to tangent and sine, respectively, and solving for x.

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106 Medium
106 of 605
Analyzing +4 | -1
If x,  y,  zx,\;y,\;z are in AP and tan⁡−1x,tan⁡−1y{\tan ^{ - 1}}x,{\tan ^{ - 1}}y and tan⁡−1z{\tan ^{ - 1}}z are also in AP, then
View answer and explanation
Correct option: A
x=y=zx = y = z
Option A is correct because if x, y, and z are in AP and their inverse tangents are also in AP, then the only solution is when x = y = z.
107 Medium
107 of 605
Application +4 | -1
If f(x)=ecosecos⁡−1x2+tan⁡−1cot⁡−1x2f(x)=\mathrm{e}^{\operatorname{cosecos}^{-1} x^2+\tan ^{-1} \cot ^{-1} x^2} then minimum value of f(x)f(x) is
View answer and explanation
Correct option: B
e2e^2
Option B (e2e^2) is correct because the minimum value of the function is obtained when x2+1x2x^2 + \frac{1}{x^2} is minimized, which occurs at x2=1x^2 = 1.
108 Medium
108 of 605
Analyzing +4 | -1
If y=tan⁡−1(log⁡e(e/x2)log⁡e(ex2))+tan⁡−1(3+2log⁡ex1−6log⁡ex),then  d2ydx2  isy = {\tan ^{ - 1}}\left( {\frac{{{{\log }_e}\left( {e/{x^2}} \right)}}{{{{\log }_e}\left( {e{x^2}} \right)}}} \right) + {\tan ^{ - 1}}\left( {\frac{{3 + 2{{\log }_e}x}}{{1 - 6{{\log }_e}x}}} \right),{\text{then}}\,\,\frac{{{d^2}y}}{{d{x^2}}}\,\,{\text{is}}
View answer and explanation
Correct option: C
0
Option D ({0}) is correct because the expression inside the square root must be non-negative, which implies that {x} must be 0.
109 Medium
109 of 605
Application +4 | -1
If cot⁡−1(1a)+cot⁡−1(1b)+cot⁡−1(1c)=π2\cot ^{-1}\left(\frac{1}{a}\right)+\cot ^{-1}\left(\frac{1}{b}\right)+\cot ^{-1}\left(\frac{1}{c}\right)=\frac{\pi}{2}, then
View answer and explanation
Correct option: B
ab+bc+ca=1a b+b c+c a=1
Option B (ab+bc+ca=1a b+b c+c a=1) is correct because it is derived from the tangent addition formula and the given condition.
110 Medium
110 of 605
Analyzing +4 | -1
If a,  ba,\;b are positive quantities and if a1=a+b2,  b1=a1b,  a2=a1+b12,  b2=a2b1{a_1} = \frac{{a + b}}{2},\;{b_1} = \sqrt {{a_1}b} ,\;{a_2} = \frac{{{a_1} + {b_1}}}{2},\;{b_2} = \sqrt {{a_2}{b_1}} and so on, then
View answer and explanation
Correct option: B
b∞=b2−a2cos⁡−1(ab){b_\infty } = \frac{{\sqrt {{b^2} - {a^2}} }}{{{{\cos }^{ - 1}}\left( {\frac{a}{b}} \right)}}
Option D is incorrect. The correct answer is b∞=b2−a2cos⁡−1(ab){b_\infty } = \frac{{\sqrt {{{{b^2} - {a^2}}}} }}{{{{\cos }^{ - 1}}\left( {\frac{a}{b}} \right)}} which is not listed, and the provided solution derives this result using trigonometric substitution and limit evaluation.
111 Hard
111 of 605
Analyzing +4 | -1
Let s={x∈R:0<x<1s=\left\{x \in \mathrm{R}: 0 < x < 1\right. and 2tan⁡−1(1−x1+x)=cos⁡−1(1−x21+x2)}\left.2 \tan ^{-1}\left(\frac{1-x}{1+x}\right)=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right\} If n(S)n(S) denotes the number of elements in SS then:
View answer and explanation
Correct option: C
n(S)=1n(S)=1 and the element in SS is less than 12\frac{1}{2}
Option C is correct because solving the given equation using trigonometric identities yields a single solution for x, which is less than 1/2.
112 Easy
112 of 605
Application +4 | -1
The principal value of tan⁡−1(cot⁡43π/4)\tan ^{-1}(\cot 43 \pi / 4) is
View answer and explanation
Correct option: D
−π/4-\pi / 4
Option D (−π/4- \pi / 4) is correct because by simplifying the expression using trigonometric identities, we find the principal value of the inverse tangent.
113 Medium
113 of 605
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 the sum of angles A and B is 3π\pi/4, making angle C equal to π\pi - 3π\pi/4 = π\pi/4.
114 Easy
114 of 605
Application +4 | -1
If sin⁡−1(x/5)+cosec⁡−1(5/4)=π/2\sin ^{-1}(x / 5)+\operatorname{cosec}^{-1}(5 / 4)=\pi / 2 then a values of xx is
View answer and explanation
Correct option: B
3
Option B (3) is correct because using the identity cosec⁡−1(x)=sin⁡−1(1/x)\operatorname{cosec}^{-1}(x) = \sin^{-1}(1/x) and sin⁡−1x+cos⁡−1x=π/2\sin^{-1}x + \cos^{-1}x = \pi/2, we can solve for xx.
115 Easy
115 of 605
Application +4 | -1
tan⁡−1(14)+tan⁡−1(29)={\tan ^{ - 1}}\left( {\frac{1}{4}} \right) + {\tan ^{ - 1}}\left( {\frac{2}{9}} \right) =
View answer and explanation
Correct option: D
tan⁡−1(12){\tan ^{ - 1}}\left( {\frac{1}{2}} \right)
Option D (tan⁡−1(12){\tan ^{ - 1}}\left( {\frac{1}{2}} \right)) is correct because it is the simplified value of the given expression after applying the tangent addition formula.
116 Easy
116 of 605
Application +4 | -1
If x∈[−1,  1]x \in \left[ { - 1,\;1} \right], then sin⁡−1(2x1+x2){\sin ^{ - 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 by substituting x=tan⁡θx = \tan \theta, the expression simplifies to sin⁡−1(sin⁡2θ){\sin ^{ - 1}}(\sin 2\theta) which equals 2θ2\theta under the given condition.
117 Easy
117 of 605
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 solving the given equation using trigonometric identities leads to x=abx = \sqrt{ab}.
118 Medium
118 of 605
Application +4 | -1
cos⁡−1[cos⁡(−17π5)]\cos ^{-1}\left[\cos \left(\frac{-17 \pi}{5}\right)\right] is equal to
View answer and explanation
Correct option: B
3π5\frac{3 \pi}{5}
Option B (3π5\frac{{3 \pi}}{{5}}) is correct because by applying the properties of cosine and inverse cosine functions, the expression simplifies to 3π5\frac{{3\pi}}{{5}}.
119 Easy
119 of 605
Application +4 | -1
The 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 equal to
View answer and explanation
Correct option: D
π3\frac{\pi}{3}
Option D (π3\frac{\pi}{3}) is correct because it is the simplified result after evaluating the inverse trigonometric functions and subtracting.
120 Easy
120 of 605
Application +4 | -1
Two angles of a triangle are cot⁡−12\cot ^{-1} 2 and cot⁡−13\cot ^{-1} 3, then the third angle are -
View answer and explanation
Correct option: B
3π4\frac{3 \pi}{4}
Option B (3π4\frac{3 \pi}{4}) is correct because the sum of the two given angles is π4\frac{\pi}{4}, and the sum of all angles in a triangle is π\pi, so the third angle is π−π4=3π4\pi - \frac{\pi}{4} = \frac{3 \pi}{4}.
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