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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Easy

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Showing 211-240 of 365 questions

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211 Easy
211 of 365
Application +4 | -1
2tan⁡−1(13)+tan⁡−1(17)2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right) is equal to
View answer and explanation
Correct option: D
π4\frac{\pi }{4}
Option D (π4\frac{\pi}{4}) is correct because using the arctangent addition formula twice, the expression simplifies to π4\frac{\pi}{4}.
212 Easy
212 of 365
Application +4 | -1
The principle value of sin⁡−1tan⁡(−5π4){\sin ^{ - 1}}\tan \left( {\frac{{ - 5\pi }}{4}} \right) is
View answer and explanation
Correct option: D
−π2- \frac{\pi }{2}
Option D (−π2- \frac{\pi }{{2}}) is correct because the expression simplifies to sin⁡−1(−1){\sin ^{ - 1}}(-1), which equals −π2- \frac{\pi }{{2}}.
213 Easy
213 of 365
Application +4 | -1
If sin⁡−1(12)+cosec⁡−1(2)\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)+\operatorname{cosec}^{-1}(2) has the value kπ12{k} \frac{\pi}{12}, then value of kk is
View answer and explanation
Correct option: A
5
Option A (5) is correct because by evaluating the inverse trigonometric functions and simplifying, the value of k is found to be 5.
214 Easy
214 of 365
Application +4 | -1
The value of xx for which sin⁡[cot⁡−1(1+x)]=cos⁡(tan⁡−1x)\sin \left[ {{{\cot }^{ - 1}}\left( {1 + x} \right)} \right] = \cos \left( {{{\tan }^{ - 1}}x} \right) is
View answer and explanation
Correct option: D
−12- \frac{1}{2}
Option D (-1/2) is correct because it is the solution obtained by simplifying the given equation using trigonometric identities and solving for x.
215 Easy
215 of 365
Application +4 | -1
The value of sin⁡−1(cos⁡2)−cos⁡−1(sin⁡2)+tan⁡−1(cot⁡4)−cot⁡−1(tan⁡4)\sin ^{-1}(\cos 2)-\cos ^{-1}(\sin 2)+\tan ^{-1}(\cot 4)-\cot ^{-1}(\tan 4) +sec⁡−1(cosec⁡6)−cosec⁡−1(sec⁡6)+\sec ^{-1}(\operatorname{cosec} 6)-\operatorname{cosec}^{-1}(\sec 6) is
View answer and explanation
Correct option: D
5π−165 \pi-16
Option D (5π−165 \pi-16) is correct because it is the simplified value of the given expression involving inverse trigonometric functions.
216 Easy
216 of 365
Application +4 | -1
Value of sin⁡(sin⁡−1(3/5)+cos⁡−1(4/5))\sin \left(\sin ^{-1}(3 / 5)+\cos ^{-1}(4 / 5)\right) is equal to
View answer and explanation
Correct option: B
24/2524 / 25
Option B (24/2524/25) is correct because it results from applying the sine addition formula and simplifying the expression.
217 Easy
217 of 365
Application +4 | -1
If α=cos⁡−1(35),β=tan⁡−1(13)\alpha=\cos ^{-1}\left(\frac{3}{5}\right), \beta=\tan ^{-1}\left(\frac{1}{3}\right), where 0<α,β<π20<\alpha, \beta<\frac{\pi}{2}, then α−β\alpha-\beta is equal to :
View answer and explanation
Correct option: D
sin⁡−1(9510)\sin ^{-1}\left(\frac{9}{5 \sqrt{10}}\right)
The correct answer is D (sin⁡−1(9510)\sin^{-1}(\frac{9}{5 \sqrt{10}})) because it is obtained by using the tangent subtraction formula and then converting to arcsin.
218 Easy
218 of 365
Application +4 | -1
If x=2/3x=2 / 3, then sin⁡2(tan⁡−1x)+cos⁡2(sin⁡−1x)\sin ^2\left(\tan ^{-1} x\right)+\cos ^2\left(\sin ^{-1} x\right) is equal to
View answer and explanation
Correct option: C
101/117101 / 117
Option C (101/117101 / 117) is correct because by substituting x=2/3x=2/3 and simplifying the expression, we get the answer 101/117101/117.
219 Easy
219 of 365
Application +4 | -1
If 0<x<120 < x < \frac{1}{{\sqrt 2 }} and sin⁡−1xα=cos⁡−1xβ\frac{{{{\sin }^{ - 1}}x}}{\alpha } = \frac{{{{\cos }^{ - 1}}x}}{\beta } then a value of sin⁡(2παα+β)\sin \left( {\frac{{2\pi \alpha }}{{\alpha + \beta }}} \right) is
View answer and explanation
Correct option: B
4x(1−x2)(1−2x2)4 x \sqrt{\left(1-x^2\right)}\left(1-2 x^2\right)
Option B is correct because by using trigonometric identities and the given condition, we can simplify the expression and find the value of sin⁡(2παα+β)\sin \left( {\frac{{2\pi \alpha }}{{\alpha + \beta }}} \right).
220 Easy
220 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 by using the double angle formula for the inverse tangent function and then the tangent subtraction formula, we can simplify the expression to tan⁡−11{\tan ^{ - 1}}1, which equals π4\frac{\pi }{4}.
221 Easy
221 of 365
Application +4 | -1
The principal value of tan⁡−1(cot⁡43π4)\tan ^{-1}\left(\cot \frac{43 \pi}{4}\right) is:
View answer and explanation
Correct option: C
−π4-\frac{\pi}{4}
The correct answer is C (−π4\frac{-\pi}{4}) because it involves simplifying the cotangent and using the properties of inverse trigonometric functions.
222 Easy
222 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 is correct because the expression simplifies using trigonometric identities and inverse trigonometric functions to x2+1x2+2\sqrt {\frac{{{x^2} + 1}}{{{x^2} + 2}}}.
223 Easy
223 of 365
Application +4 | -1
If x=sin⁡(2tan⁡−12)x = \sin \left( {2{{\tan }^{ - 1}}2} \right) and y=sin⁡(12tan⁡−143)y = \sin \left( {\frac{1}{2}{{\tan }^{ - 1}}\frac{4}{3}} \right), then
View answer and explanation
Correct option: B
y2=1−x{y^2} = 1 - x
Option B (y2=1−x{y^2} = 1 - x) is correct because it represents the relationship between x and y after simplifying the given trigonometric expressions.
224 Easy
224 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 tangent subtraction formula, we can simplify the expression and find that A - B equals 30 degrees.
225 Easy
225 of 365
Analyzing +4 | -1
2π−(sin⁡−145+sin⁡−1513+sin⁡−11665)2 \pi-\left(\sin ^{-1} \frac{4}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{16}{65}\right) is equal to :
View answer and explanation
Correct option: C
3π2\frac{3 \pi}{2}
Option C (3π2\frac{{3 \pi}}{{2}}) is correct because by converting the inverse sines to inverse tangents and using the tangent addition formula, the expression simplifies to 2π−π2=3π22\pi - \frac{\pi}{2} = \frac{{3\pi}}{{2}}.

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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
226 Easy
226 of 365
Application +4 | -1
Sin⁡(2sin⁡−10.8)=\operatorname{Sin}\left(2 \sin ^{-1} 0.8\right)=
View answer and explanation
Correct option: A
0.96
Option A (0.96) is correct because it is the result of applying the double angle formula for sine and simplifying the expression.
227 Easy
227 of 365
Application +4 | -1
tan⁡−1xa2−x2\tan ^{-1} \frac{x}{\sqrt{a^2-x^2}} is equal to
View answer and explanation
Correct option: C
sin⁡−1(xa)\sin ^{-1}\left(\frac{x}{a}\right)
Option C (sin⁡−1(xa)\sin ^{-1}(\frac{{x}}{{a}})) is correct because by substituting x=asin⁡θx = a \sin \theta, the expression simplifies to this inverse sine function.
228 Easy
228 of 365
Analyzing +4 | -1
If xx takes negative permissible value, then sin⁡−1x{\sin ^{ - 1}}x is equal to
View answer and explanation
Correct option: A
−cos⁡−11−x2- {\cos ^{ - 1}}\sqrt {1 - {x^2}}
Option A is correct because it uses the property that sin-1(-x) = -sin-1(x) and the relationship between sin-1(x) and cos-1(sqrt(1-x2)) when x is negative.
229 Easy
229 of 365
Application +4 | -1
cot⁡−1[(cos⁡α)1/2]−tan⁡−1[(cos⁡α)1/2]=x\cot ^{-1}\left[(\cos \alpha)^{1 / 2}\right]-\tan ^{-1}\left[(\cos \alpha)^{1 / 2}\right]=x then sin⁡x=\sin x=
View answer and explanation
Correct option: A
tan⁡2(α/2)\tan ^2(\alpha / 2)
Option A (tan⁡2(α/2)\tan ^2(\alpha / 2)) is correct because by simplifying the given equation using trigonometric identities, we find that sin⁡x=tan⁡2(α/2)\sin x = \tan^2(\alpha / 2).
230 Easy
230 of 365
Application +4 | -1
If x2+y2+z2=r2,{x^2} + {y^2} + {z^2} = {r^2},, then

tan⁡−1(xyzr)+tan⁡−1(yzxr)+tan⁡−1(xzyr)  {\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{xz}}{{yr}}} \right)\; is equal to
View answer and explanation
Correct option: B
π2\frac{\pi }{2}
Option B (π2\frac{\pi}{2}) is correct because the expression simplifies to tan⁡−1(∞)\tan^{-1}(\infty) using the given condition x2+y2+z2=r2{x^2} + {y^2} + {z^2} = {r^2}.
231 Easy
231 of 365
Application +4 | -1
tan⁡−1(2x1−x2)+cot⁡−1(1−x22x)=π3,x∈[−1,1]\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{3}, x \in[-1,1] sum of all solutions is α−43\alpha-\frac{4}{\sqrt{3}}, then α\alpha is:
View answer and explanation
Correct option: B
2
Option B (2) is correct because after solving the equation, the sum of all solutions is indeed 2 - 4/sqrt(3), making alpha equal to 2.
232 Easy
232 of 365
Application +4 | -1
The value of sin⁡[π2−sin⁡−1  (−32)]\sin \left[ {\frac{\pi }{2} - {{\sin }^{ - 1}}\;\left( { - \frac{{\sqrt 3 }}{2}} \right)} \right] is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (1/2) is correct because simplifying the expression using trigonometric identities and inverse trigonometric functions leads to this result.
233 Easy
233 of 365
Application +4 | -1
If tan⁡−111+2+tan⁡−111+(2)(3)+tan⁡−111+(3)(4)+…tan⁡−111+n(n+1)=tan⁡−1θ\tan ^{-1} \frac{1}{1+2}+\tan ^{-1} \frac{1}{1+(2)(3)}+\tan ^{-1} \frac{1}{1+(3)(4)}+\ldots \tan ^{-1} \frac{1}{1+n(n+1)}=\tan ^{-1} \theta, then θ=\theta=
View answer and explanation
Correct option: C
nn+2\frac{n}{n+2}
Option C (nn+2\frac{n}{n+2}) is correct because the given series telescopes to tan⁡−1(n+1)−tan⁡−1(1)\tan^{-1}(n+1) - \tan^{-1}(1), which simplifies to tan⁡−1(nn+2)\tan^{-1}(\frac{n}{n+2}).
234 Easy
234 of 365
Application +4 | -1
If tan⁡−1(1/4)+tan⁡−1(2/9)=(1/2)cos⁡−1x\tan ^{-1}(1 / 4)+\tan ^{-1}(2 / 9)=(1 / 2) \cos ^{-1} x then xx is equal to
View answer and explanation
Correct option: C
3/53 / 5
Option C (3/5) is correct because, after simplifying the left side using the tangent addition formula and then using the double angle formula for cosine, we find that x = 3/5.
235 Easy
235 of 365
Remembering +4 | -1
Which of the following is the principal value branch of cosec⁡−1x\operatorname{cosec}^{-1} \mathrm{x} ?
View answer and explanation
Correct option: D
[−π2,π2]−{0}\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}
Option D ([−π2,π2]−{0}\left[\frac{{-\pi}}{{2}}, \frac{{\pi}}{{2}}\right] - \{0\}) is correct because the principal value branch of cosec⁡−1x\operatorname{cosec}^{-1} x is [−π2,π2]−{0}\left[\frac{{-\pi}}{{2}}, \frac{{\pi}}{{2}}\right] - \{0\}.
236 Easy
236 of 365
Analyzing +4 | -1
Let  SK=∑r=1Ktan⁡−1(6r22r+1+32r+1)\,{S_K} = \sum\limits_{r = 1}^K {{{\tan }^{ - 1}}\left( {\frac{{{6^r}}}{{{2^{2r + 1}} + {3^{2r + 1}}}}} \right)}. Then  lim⁡   K→∞ SK\,\mathop {\lim \,\,\,}\limits_{K \to \infty } \,{S_K} is equal to:
View answer and explanation
Correct option: B
cot⁡−1(32){\cot ^{ - 1}}\left( {\frac{3}{2}} \right)
Option B (cot⁡−1(32){\cot ^{ - 1}}\left( {\frac{3}{2}} \right)) is correct because the summation telescopes, and the limit is found using inverse trigonometric identities.
237 Easy
237 of 365
Analyzing +4 | -1
The number of roots of the equation sin⁡−1x+sin⁡−11−x3=π2\sin ^{-1} \sqrt{x}+\sin ^{-1} \sqrt{1-x^3}=\frac{\pi}{2} is equal to
View answer and explanation
Correct option: C
2
The correct answer is C (2) because solving the equation sin⁡−1x+sin⁡−11−x3=π2\sin^{-1} \sqrt{x} + \sin^{-1} \sqrt{1-x^3} = \frac{\pi}{2} yields two valid solutions: x = 0 and x = 1.
238 Easy
238 of 365
Application +4 | -1
If ∑r=120tan⁡−1(12r2)=tan⁡−1mn\sum_{\mathrm{r}=1}^{20} \tan ^{-1}\left(\frac{1}{2 \mathrm{r}^2}\right)=\tan ^{-1} \frac{\mathrm{m}}{\mathrm{n}}; where m\mathrm{m} and n\mathrm{n} are co-prime numbers then [log⁡2(m+n)]\left[\log _2(m+n)\right] equals (where [.] is greatest integer function)
View answer and explanation
Correct option: A
5
Option A (15) is correct because cot⁡71∘2=6+4+3+2\cot 7 \frac{1^{\circ}}{2} = \sqrt{6} + \sqrt{4} + \sqrt{3} + \sqrt{2}, and the sum of the numbers under the square roots is 15.
239 Easy
239 of 365
Application +4 | -1
The value of sin⁡−1(cos⁡33π5){\sin ^{ - 1}}\left( {\cos \frac{{33\pi }}{5}} \right) is
View answer and explanation
Correct option: D
−π10- \frac{\pi }{{10}}
Option D (−π10- \frac{\pi }{{10}}) is correct because simplifying the expression sin⁡−1(cos⁡33π5){\sin ^{ - 1}}\left( {\cos \frac{{33\pi }}{5}} \right) leads to −π10- \frac{\pi }{{10}}.
240 Easy
240 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 (32\frac{{\sqrt 3 }}{2}) is correct because by using the identity sin⁡−1x+cos⁡−1x=π2{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}, we can simplify the equation and solve for x.
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