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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Medium

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Showing 181-210 of 221 questions
181 Medium
181 of 221
Analyzing +4 | -1
S=tan⁡−1(1n2+n+1)+tan⁡−1(1n2+3n+3)+…S=\tan ^{-1}\left(\frac{1}{n^{2}+n+1}\right)+\tan ^{-1}\left(\frac{1}{n^{2}+3 n+3}\right)+\ldots
+tan⁡−1(11+(n+19)(n+20))+\tan ^{-1}\left(\frac{1}{1+(n+19)(n+20)}\right), then tan⁡S\tan S is equal to :
View answer and explanation
Correct option: C
20n2+20n+1\frac{20}{n^{2}+20 n+1}
The correct answer is C (20n2+20n+1\frac{20}{n^{2}+20 n+1}). This is found by recognizing the telescoping nature of the series and using the arctangent subtraction formula.
182 Medium
182 of 221
Application +4 | -1
If 12≤x≤1\frac{1}{2} \leq x \leq 1 then sin⁡−1(3x−4x3)\sin ^{-1}\left(3 x-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 when 12≤x≤1\frac{1}{2} \leq x \leq 1, the range of 3sin⁡−1x3 \sin^{-1} x is such that sin⁡−1(3x−4x3)\sin^{-1}(3x - 4x^3) needs to be expressed as π−3sin⁡−1x\pi - 3 \sin^{-1} x to fall within the principal range of sin⁡−1\sin^{-1}.
183 Medium
183 of 221
Application +4 | -1
If the matrix A=[1−11−2]A=\left[\begin{array}{rr}1 & -1 \\ 1 & -2\end{array}\right] satisfies the equation An=5I−8AA^n=5 I-8 A, then nn is equal to
View answer and explanation
Correct option: A
6
Option A (6) is correct because by finding a pattern in the powers of A, we can determine that A6=5I−8AA^6 = 5I - 8A.
184 Medium
184 of 221
Application +4 | -1
If sin⁡−113+sin⁡−123=sin⁡−1x\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{2}{3}=\sin ^{-1} x, then xx is equal to:
View answer and explanation
Correct option: C
5+429\frac{\sqrt{5}+4 \sqrt{2}}{9}
Option C (5+429\frac{{\sqrt{{5}}+4 \sqrt{{2}}}}{{9}}) is correct because applying the sine addition formula and simplifying results in this value for x.
185 Medium
185 of 221
Application +4 | -1
The principal value of sin⁡−1(sin⁡10)\sin ^{-1}(\sin 10) is
View answer and explanation
Correct option: C
3π−103 \pi-10
Option C (3π−103\pi - 10) is correct because the principal value of sin⁡−1(sin⁡x)\sin^{-1}(\sin x) must lie in the range [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}], and 3π−103\pi - 10 falls within this range.
186 Medium
186 of 221
Application +4 | -1
The value of cos⁡{tan⁡−1(tan⁡2)}\cos \left\{ {{{\tan }^{ - 1}}(\tan 2)} \right\}, is
View answer and explanation
Correct option: D
−cos⁡2- \cos 2
Option D (−cos⁡2- \cos 2) is correct because by using the property of inverse trigonometric functions, the expression simplifies to −cos⁡2- \cos 2.
187 Medium
187 of 221
Application +4 | -1
cos⁡−1(1517)+2tan⁡−1(15)\cos ^{-1}\left(\frac{15}{17}\right)+2 \tan ^{-1}\left(\frac{1}{5}\right) is equal to
View answer and explanation
Correct option: B
sin⁡−1(171221)\sin ^{-1}\left(\frac{171}{221}\right)
Option B (sin⁡−1(171221)\sin^{-1}(\frac{{171}}{{221}})) is correct because simplifying the expression using trigonometric identities results in this value.
188 Medium
188 of 221
Analyzing +4 | -1
cot⁡−1(cos⁡α)−tan⁡−1(cos⁡α)=x\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x, then sin⁡x=\sin x=
View answer and explanation
Correct option: A
tan⁡2(α2)\tan ^{2}\left(\frac{\alpha}{2}\right)
Option A (tan⁡2(α2)\tan ^{2}\left(\frac{{\alpha}}{{2}}\right)) is correct because by using trigonometric identities and simplifying the expression, we find that sin⁡x=tan⁡2(α2)\sin x = \tan^2(\frac{\alpha}{2}).
189 Medium
189 of 221
Application +4 | -1
If cos⁡−1x−cos⁡−1y2=α\cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha, where −1≤x≤1,−2≤y≤2-1 \leq x \leq 1,-2 \leq y \leq 2, x≤y2x \leq \frac{y}{2}, then for all x,y,4x2−4xycos⁡α+y2x, y, 4 x^{2}-4 x y \cos \alpha+y^{2} is equal to:
View answer and explanation
Correct option: A
4sin⁡2α4 \sin ^{2} \alpha
The correct answer is A (4sin⁡2α4 \sin^2 \alpha) because it involves using the cosine subtraction formula and trigonometric identities to simplify the expression.
190 Medium
190 of 221
Application +4 | -1
If cos⁡−1p+cos⁡−1q+cos⁡−1r=π\cos ^{-1} \mathrm{p}+\cos ^{-1} \mathrm{q}+\cos ^{-1} \mathrm{r}=\pi. Then value of p2+q2\mathrm{p}^2+\mathrm{q}^2 +r2+2pqr+4+r^2+2 p q r+4 is equal to-
View answer and explanation
Correct option: C
5
Option C (5) is correct because the given condition leads to the identity p2+q2+r2+2pqr=1p^2 + q^2 + r^2 + 2pqr = 1, and adding 4 to both sides results in 5.
191 Medium
191 of 221
Analyzing +4 | -1
sin⁡−1(sin⁡3x2−12x2+2)<34\sin ^{-1}\left(\sin \frac{3 x^2-1}{2 x^2+2}\right) < \frac{3}{4} if
View answer and explanation
Correct option: A
−53<x<53-\sqrt{\frac{5}{3}} < x < \sqrt{\frac{5}{3}}
Option C (1/e) is correct because by applying L'Hopital's rule to the logarithm of the expression, we find the limit to be -1, thus the original limit is 1/e.
192 Medium
192 of 221
Application +4 | -1
sec⁡2(tan⁡−12)+cosec⁡2(cot⁡−13)\sec ^2\left(\tan ^{-1} 2\right)+\operatorname{cosec}^2\left(\cot ^{-1} 3\right) is equal to
View answer and explanation
Correct option: D
15
Option D (15) is correct because sec⁡2(tan⁡−12)+cosec⁡2(cot⁡−13)=(1+tan⁡2(tan⁡−12))+(1+cot⁡2(cot⁡−13))=1+4+1+9=15\sec^2(\tan^{-1} 2) + \operatorname{cosec}^2(\cot^{-1} 3) = (1 + \tan^2(\tan^{-1} 2)) + (1 + \cot^2(\cot^{-1} 3)) = 1 + 4 + 1 + 9 = 15.
193 Medium
193 of 221
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 the expression simplifies using trigonometric identities to this value when x = 1/5.
194 Medium
194 of 221
Application +4 | -1
cos⁡[2tan⁡−1(17)]\cos \left[2 \tan ^{-1}\left(\frac{1}{7}\right)\right] equals
View answer and explanation
Correct option: A
sin⁡(4cot⁡−13)\sin \left(4 \cot ^{-1} 3\right)
Option A (sin⁡(4cot⁡−13)\sin \left(4 \cot ^{-1} 3\right)) is correct because by applying trigonometric identities, both expressions simplify to 2425\frac{{24}}{{25}}.
195 Medium
195 of 221
Application +4 | -1
Let tan⁡−1y=tan⁡−1x+tan⁡−1(2x1−x2)\tan ^{-1} y=\tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right), where or ∣x∣<13|x|<\frac{1}{\sqrt{3}}. Then a value of yy is :
View answer and explanation
Correct option: A
3x−x31−3x2\frac{3 x-x^{3}}{1-3 x^{2}}
The correct answer is A (3x−x31−3x2\frac{3 x-x^{3}}{1-3 x^{2}}) because it involves using the triple angle formula for tangent.

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196 Medium
196 of 221
Application +4 | -1
If x∈(1,  ∞)x \in \left( {1,\;\infty } \right), then sin⁡−1(2x1+x2){\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right) equals
View answer and explanation
Correct option: B
π−2tan⁡−1x\pi - 2{\tan ^{ - 1}}x
Option B (π−2tan⁡−1x\pi - 2{\tan ^{ - 1}}x) is correct because when x∈(1,∞)x \in (1, \infty), the range of 2tan⁡−1x2\tan^{-1}x is (π2,π)(\frac{\pi}{2}, \pi), and we need to adjust the result of sin⁡−1(sin⁡(2θ))\sin^{-1}(\sin(2\theta)) to fall within the principal range of sin⁡−1\sin^{-1}.
197 Medium
197 of 221
Understanding +4 | -1
tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy)\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) for
View answer and explanation
Correct option: D
xy<1x y<1
Option D (xy<1xy<1) is correct because the formula for the sum of arctangents is valid only when the product of the arguments is less than 1.
198 Medium
198 of 221
Application +4 | -1
Given that sin⁡−1(sin⁡3π4)=2πk\sin ^{-1}\left(\sin \frac{3 \pi}{4}\right)=\frac{2 \pi}{\mathrm{k}}, then k={k}=
View answer and explanation
Correct option: A
8
Option A (8) is correct because by simplifying the expression using trigonometric properties, the value of k is found to be 8.
199 Medium
199 of 221
Application +4 | -1
The value of sin⁡−1[cos⁡{cos⁡−1(cos⁡x)+sin⁡−1(sin⁡x)}]\sin ^{-1}\left[\cos \left\{\cos ^{-1}(\cos x)+\sin ^{-1}(\sin x)\right\}\right]; where x∈(π2,π)x \in\left(\frac{\pi}{2}, \pi\right) is
View answer and explanation
Correct option: D
−π2\frac{-\pi}{2}
Option D (−π2\frac{-\pi}{2}) is correct because, for x∈(π2,π)x \in (\frac{\pi}{2}, \pi), sin⁡−1(sin⁡x)=π−x\sin^{-1}(\sin x) = \pi - x and cos⁡−1(cos⁡x)=x\cos^{-1}(\cos x) = x, leading to sin⁡−1(cos⁡(π))=−π2\sin^{-1}(\cos(\pi)) = \frac{-\pi}{2}.
200 Medium
200 of 221
Application +4 | -1
If (sin⁡−1x)2−(cos⁡−1x)2=a;0<x<1;a≠0{\left( {{{\sin }^{ - 1}}x} \right)^2} - {\left( {{{\cos }^{ - 1}}x} \right)^2} = a;0 < x < 1;a \ne 0 , then the value of 2x2−12{x^2} - 1 is
View answer and explanation
Correct option: B
sin⁡(2aπ)\sin \left(\frac{2 a}{\pi}\right)
Option B (sin(2a/π)) is correct because using the identity sin-1(x) + cos-1(x) = π/2, the given equation can be simplified to express cos-1(x) in terms of a, and then 2x2 - 1 can be expressed as sin(2a/π).
201 Medium
201 of 221
Application +4 | -1
For which value of x,sin⁡[cot⁡−1(x+1)]=cos⁡(tan⁡−1x)x, \sin \left[\cot ^{-1}(x+1)\right]=\cos \left(\tan ^{-1} x\right) is
View answer and explanation
Correct option: D
−12-\frac{1}{2}
Option D (−12-\frac{1}{2}) is correct because it is the solution to the equation obtained by equating the sine and cosine expressions after simplification.
202 Medium
202 of 221
Application +4 | -1
The number of positive integral solutions of the equation tan⁡−1x+cot⁡−1y=tan⁡−13\tan ^{-1} x+\cot ^{-1} y=\tan ^{-1} 3, is
View answer and explanation
Correct option: A
two
Option A (two) is correct because there are only two positive integral solutions that satisfy the given equation.
203 Medium
203 of 221
Analyzing +4 | -1
If the sum of the series tan⁡−13+tan⁡−192+tan⁡−1334+……\tan ^{-1} 3+\tan ^{-1} \frac{9}{2}+\tan ^{-1} \frac{33}{4}+\ldots \ldots to nn terms == (2n+1)πA+Btan⁡−1(2n)\frac{(2 n+1) \pi}{A}+B \tan ^{-1}\left(2^n\right) then (A+B)(A+B) is equal to
View answer and explanation
Correct option: D
3
Option A (-√(5/3) < x < √(5/3)) is correct because solving the inequality leads to the condition x² < 5/3, which gives the specified range for x.
204 Medium
204 of 221
Application +4 | -1
If f(x)=2tan⁡−1x+sin⁡−1(2x1+x2),x>1f(x)=2 \tan ^{-1} x+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right), x>1 then f(5)f(5) is equal to :
View answer and explanation
Correct option: C
π\pi
Option C (π\pi) is correct because after simplification, the function f(x)f(x) becomes a constant function equal to π\pi for x>1x>1.
205 Medium
205 of 221
Application +4 | -1
The value of cos⁡(2cos⁡−1x+sin⁡−1x)\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right) at x=15x=\frac{1}{5} is
View answer and explanation
Correct option: A
−265-\frac{2 \sqrt{6}}{5}
Option A (−265-\frac{2 \sqrt{6}}{5}) is correct because the expression simplifies using trigonometric identities and the given value of x.
206 Medium
206 of 221
Application +4 | -1
If tan⁡−1x−1x−2+tan⁡−1x+1x+2=π4\tan ^{-1} \frac{x-1}{x-2}+\tan ^{-1} \frac{x+1}{x+2}=\frac{\pi}{4}, then x=x=
View answer and explanation
Correct option: C
±12\pm \frac{1}{\sqrt{2}}
Option C (±12\pm \frac{{1}}{{\sqrt{{2}}}}) is correct because it is obtained by applying the tangent addition formula and solving the resulting quadratic equation.
207 Medium
207 of 221
Application +4 | -1
Solution of the equation 2tan⁡−1(cos⁡x)=tan⁡−1(2cosec⁡x)2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x) is
View answer and explanation
Correct option: B
x=nπ+π4;n∈Zx=n \pi+\frac{\pi}{4} ; n \in \mathbb{Z}
Option B (x=nπ+π4;n∈Zx=n \pi+\frac{\pi}{4} ; n \in \mathbb{Z}) is correct because it represents the solution to the given equation after applying trigonometric identities and solving for x.
208 Medium
208 of 221
Analyzing +4 | -1
If f(x)=sin⁡−1(x−x−1x(x+1))f(x)=\sin ^{-1}\left(\frac{\sqrt{x}-\sqrt{x-1}}{\sqrt{x(x+1)}}\right) then f1(x)=f^1(x)=
View answer and explanation
Correct option: A
12(1x(x+1)−1xx−1)\frac{1}{2}\left(\frac{1}{\sqrt{x}(x+1)}-\frac{1}{x \sqrt{x-1}}\right)
Option A is correct. The derivative is found by first simplifying the function using trigonometric identities and then differentiating.
209 Medium
209 of 221
Application +4 | -1
The value of cos⁡−1(cot⁡(sin⁡−1(2−34)+cos⁡−1(124)+sec⁡−1(2)))\cos ^{-1}\left(\cot \left(\sin ^{-1}\left(\sqrt{\frac{2-\sqrt{3}}{4}}\right)+\cos ^{-1}\left(\frac{\sqrt{12}}{4}\right)+\sec ^{-1}(\sqrt{2})\right)\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 cos⁡−1(0)\cos^{-1}(0), which equals π2\frac{\pi}{2}.
210 Medium
210 of 221
Application +4 | -1
If cos⁡−1x+cos⁡−1y+cos⁡−1z=π\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\pi, then
View answer and explanation
Correct option: B
x2+y2+z2+2xyz=1x^2+y^2+z^2+2 x y z=1
Option B (x2+y2+z2+2xyz=1x^2+y^2+z^2+2xyz=1) is correct because using the identity cos⁡−1x+cos⁡−1y+cos⁡−1z=π\cos^{-1} x + \cos^{-1} y + \cos^{-1} z = \pi and simplifying the expression leads to the correct equation.
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