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JEE • MATHEMATICS • INVERSE TRIGONOMETRY

Inverse trigonometry practice · Medium

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Showing 211-221 of 221 questions
211 Medium
211 of 221
Analyzing +4 | -1
The number of solutions of the equation cos⁡−1(1−x)+mcos⁡−1(x)=nπ2\cos ^{-1}(1-x)+m \cos ^{-1}(x)=\frac{n \pi}{2}, where m>0m > 0, n≤0n \leq 0, is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the limit from both sides (x → 0+ and x → 0-) results in [sin(1)/1] = 0 and [sin(-1)/-1] = 0, respectively.
212 Medium
212 of 221
Application +4 | -1
tan⁡−1(xy)−tan⁡−1(x−yx+y)\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right) is
View answer and explanation
Correct option: D
π4\frac{\pi}{4} or −3π4-\frac{3 \pi}{4}
Option D (π4\frac{{\pi}}{{4}} or −3π4-\frac{{3 \pi}}{{4}}) is correct because simplifying the expression using tangent subtraction and properties of inverse trigonometric functions yields these two possible values.
213 Medium
213 of 221
Application +4 | -1
cos⁡[cos⁡−1(−17)+sin⁡−1(−17)]\cos \left[ {{{\cos }^{ - 1}}\left( { - \frac{1}{7}} \right) + {{\sin }^{ - 1}}\left( { - \frac{1}{7}} \right)} \right] is equal to
View answer and explanation
Correct option: B
0
Option B (0) is correct because the expression simplifies to cos(π/2), which equals 0.
214 Medium
214 of 221
Application +4 | -1
A root of the equation 17x2+17xtan⁡[2tan⁡−1(1/5)−π/4]−10=017 x^2+17 x \tan \left[2 \tan ^{-1}(1 / 5)-\pi / 4\right]-10=0 is
View answer and explanation
Correct option: D
1
Option D (1) is correct because by simplifying the expression inside the tangent function and substituting it back into the equation, we find that x=1x=1 is a root.
215 Medium
215 of 221
Application +4 | -1
Let A={1,2,3,4}A=\{1,2,3,4\} and x,y∈Ax, y \in A, then the value of max⁡(sin⁡−1(sin⁡x)+cos⁡−1(cos⁡y))\max \left(\sin ^{-1}(\sin x)+\cos ^{-1}(\cos y)\right) is
View answer and explanation
Correct option: B
π+1\pi+1
Option D ((4,25)(4,2 \sqrt{5})) is correct because the point lies on the director circle of the given circle, ensuring the tangents drawn from that point are at right angles.
216 Medium
216 of 221
Application +4 | -1
sin⁡−1sin⁡(16)\sin ^{-1} \sin (16) is equal to
View answer and explanation
Correct option: A
5π−165 \pi-16
Option A (5π−165 \pi-16) is the correct answer because it falls within the principal range of the inverse sine function after applying trigonometric identities.
217 Medium
217 of 221
Application +4 | -1
If (tan⁡−1x)2+(cos⁡−1x)2=5π28\left(\tan ^{-1} x\right)^2+\left(\cos ^{-1} x\right)^2=\frac{5 \pi^2}{8}, then xx equals
View answer and explanation
Correct option: A
−1-1
Option A (-1) is correct because it satisfies the given equation after using the identity relating tan⁡−1x\tan^{-1} x and cos⁡−1x\cos^{-1} x.
218 Medium
218 of 221
Application +4 | -1
The number of positive integral solutions of the equation tan⁡−1x+cos⁡−1y1+y2=sin⁡−1310{\tan ^{ - 1}}x + {\cos ^{ - 1}}\frac{y}{{\sqrt {1 + {y^2}} }} = {\sin ^{ - 1}}\frac{3}{{\sqrt {10} }} is
View answer and explanation
Correct option: B
Two
Option B (Two) is correct because there are two positive integral solutions (1,2) and (2,7) that satisfy the given equation after simplification and trial.
219 Medium
219 of 221
Application +4 | -1
The value of cos⁡(12cos⁡−118)\cos \left(\frac{1}{2} \cos ^{-1} \frac{1}{8}\right) is equal to
View answer and explanation
Correct option: A
3/43 / 4
Option A (3/4) is correct because using the half-angle formula for cosine, we can find the value of the expression.
220 Medium
220 of 221
Analyzing +4 | -1
Let SS be the set of all solutions of the equation
cos⁡−1(2x)−2cos⁡−1(1−x2)=π,x∈[−12,12]. \cos ^{-1}(2 x)-2 \cos ^{-1}\left(\sqrt{1-x^{2}}\right)=\pi, x \in\left[-\frac{1}{2}, \frac{1}{2}\right] \text {. }
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: A
−2π3\frac{-2 \pi}{3}
Option A (−2π3\frac{{-2 \pi}}{{3}}) is correct because solving the equation and substituting the result into the expression yields this value.
221 Medium
221 of 221
Application +4 | -1
cos⁡−1(12(cos⁡7π5−sin⁡2π5))\cos ^{-1}\left(\frac{1}{\sqrt{2}}\left(\cos \frac{7 \pi}{5}-\sin \frac{2 \pi}{5}\right)\right) is equal to
View answer and explanation
Correct option: D
17π20\frac{17 \pi}{20}
Option D (17π20\frac{17 \pi}{20}) is correct because it is obtained by simplifying the expression inside the inverse cosine function using trigonometric identities and then evaluating the inverse cosine.
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