JEE Mathematics Inverse Trigonometric Functions MCQs Set 08

Answer: (a) \( \frac{43\pi}{12} \)

Explanation:
1. \( \cos^{-1}\left(-\frac12\right)=\pi-\frac\pi3=\frac{2\pi}{3},\ \sin^{-1}\frac12=\frac\pi6,\ \cos^{-1}\left(-\frac1{\sqrt2}\right)=\frac{3\pi}{4},\ \tan^{-1}(-1)=-\frac\pi4 \).
2. \( \frac{2\pi}{3}-\frac{\pi}{3}+\frac{9\pi}{4}+\pi=\frac{8\pi-4\pi+27\pi+12\pi}{12}=\frac{43\pi}{12} \).

Teacher's Note:
1. Reduce every term to a principal value first: cos inverse of a negative number is pi minus the positive value, and tan inverse is odd.
2. Use a common denominator of 12 only at the end to avoid fraction slips.

Answer: (a) \( \frac{\sqrt5}{3} \)

Explanation:
1. \( \sec^{-1}\frac1x=\cos^{-1}x\Rightarrow\tan(\cos^{-1}x)=\frac{\sqrt{1-x^2}}{x} \).
2. \( \sin(\tan^{-1}2)=\frac{2}{\sqrt{1+4}}=\frac{2}{\sqrt5} \).
3. \( \frac{\sqrt{1-x^2}}{x}=\frac{2}{\sqrt5}\Rightarrow5(1-x^2)=4x^2\Rightarrow x^2=\frac59\Rightarrow x=\frac{\sqrt5}{3} \) (taking \( x>0 \)).

Teacher's Note:
1. Replace sec inverse of 1/x by cos inverse of x so that the right-angled triangle method applies.
2. Squaring removes the radical, so check that the root is positive and at most 1 for the inverse functions to exist.

Answer: (a) \( \frac\pi3 \)

Explanation:
1. \( \frac x2+\frac{\sqrt3}{2}\sqrt{1-x^2}=\cos\frac\pi3\cdot x+\sin\frac\pi3\sqrt{1-x^2} \).
2. With \( x=\cos\theta,\ 0\le\theta\le\frac\pi3 \) this equals \( \cos\left(\theta-\frac\pi3\right) \), and \( \theta-\frac\pi3\in\left[-\frac\pi3,0\right] \), so the first term is \( \frac\pi3-\theta \).
3. Adding \( \cos^{-1}x=\theta \) gives \( \frac\pi3 \).

Teacher's Note:
1. Spot the pattern cos A cos B + sin A sin B with A = theta and B = pi/3.
2. The condition 1/2 <= x <= 1 keeps theta - pi/3 inside [-pi/3, 0], which is within the range of cos inverse after taking absolute value.

Answer: (a) \( \frac{6}{17} \)

Explanation:
1. \( \operatorname{cosec}^{-1}\frac53=\sin^{-1}\frac35=\tan^{-1}\frac34 \).
2. \( \tan^{-1}\frac34+\tan^{-1}\frac23=\tan^{-1}\frac{\frac34+\frac23}{1-\frac12}=\tan^{-1}\frac{17}{6} \) (product \( \frac12<1 \)).
3. \( \cot\left(\tan^{-1}\frac{17}{6}\right)=\frac{6}{17} \).

Teacher's Note:
1. Convert cosec inverse to sin inverse, then to tan inverse with a 3-4-5 triangle.
2. Cotangent of tan inverse of t is simply 1/t.

Answer: (a) \( \left(\frac{1}{\sqrt2},1\right] \)

Explanation:
1. \( \sin^{-1}x-\cos^{-1}x=2\sin^{-1}x-\frac\pi2>0\Rightarrow\sin^{-1}x>\frac\pi4 \).
2. Since sin inverse is increasing, \( x>\sin\frac\pi4=\frac1{\sqrt2} \).
3. Together with the domain \( |x|\le1 \): \( x\in\left(\frac1{\sqrt2},1\right] \).

Teacher's Note:
1. Use sin inverse x + cos inverse x = pi/2 to express the inequality in a single inverse function.
2. Strict inequality means the end point 1/root 2 is excluded, where the difference is exactly zero.

Answer: (a) \( \tan^{-1}(n+1)-\frac\pi4 \)

Explanation:
1. Let \( y=\operatorname{cosec}^{-1}\sqrt{(n^2+1)(n^2+2n+2)} \). Then \( \cot^2y=(n^2+1)(n^2+2n+2)-1=(n^2+n+1)^2 \).
2. \( \tan y=\frac{1}{n^2+n+1}=\frac{(n+1)-n}{1+n(n+1)}\Rightarrow y=\tan^{-1}(n+1)-\tan^{-1}n \).
3. The sum telescopes: \( \sum y=\tan^{-1}(n+1)-\tan^{-1}1=\tan^{-1}(n+1)-\frac\pi4 \).

Teacher's Note:
1. Write the general term as a difference of two tan inverse values so that consecutive terms cancel.
2. Check with n = 1: cosec inverse of root 10 equals tan inverse (1/3), which equals tan inverse 2 minus pi/4.

Answer: (a) \( 4\pi-10 \)

Explanation:
1. Here \( 10\ \text{rad} \) is not in \( [0,\pi] \).
2. \( \cos10=\cos(4\pi-10) \) and \( 4\pi-10\approx2.57\in[0,\pi] \).
3. Hence \( \cos^{-1}(\cos10)=4\pi-10 \).

Teacher's Note:
1. cos inverse of cos theta returns the angle in [0, pi]; use cos(2 n pi - theta) = cos theta to shift into it.
2. Always check the final value is non-negative and at most pi.

Answer: (a) \( 2\pi-6 \)

Explanation:
1. \( \tan^{-1}(\tan\theta)=\theta \) only for \( -\frac\pi2<\theta<\frac\pi2 \), and \( -6 \) is outside this interval.
2. \( 2\pi-6\approx0.28\in\left(-\frac\pi2,\frac\pi2\right) \) and \( \tan(2\pi-6)=\tan(-6) \).
3. Hence \( \tan^{-1}(\tan(-6))=2\pi-6 \).

Teacher's Note:
1. tan has period pi, so you may add or subtract any multiple of pi; here adding 2 pi works because it is the nearest shift that lands in the interval.
2. The answer is small and positive because -6 rad is just below -2 pi.

Answer: (a) \( \left[-\frac{3\pi}{2},\frac\pi2\right] \)

Explanation:
1. \( \sin^{-1}x-\cos^{-1}x=\sin^{-1}x-\left(\frac\pi2-\sin^{-1}x\right)=2\sin^{-1}x-\frac\pi2 \).
2. As \( \sin^{-1}x\in\left[-\frac\pi2,\frac\pi2\right] \), the expression varies from \( -\pi-\frac\pi2=-\frac{3\pi}{2} \) to \( \pi-\frac\pi2=\frac\pi2 \).

Teacher's Note:
1. Rewrite everything in terms of one inverse function and then use its range.
2. The end values correspond to x = -1 and x = 1.

Answer: (c) C, A, B

Explanation:
1. \( \log_32\approx0.63\in(0,1)\Rightarrow A=\sin^{-1}(0.63)\in\left(0,\frac\pi2\right) \).
2. \( \log_3\frac12\approx-0.63\Rightarrow B=\cos^{-1}(-0.63)\in\left(\frac\pi2,\pi\right) \).
3. \( \log_{1/3}2=-\log_32\approx-0.63\Rightarrow C=\tan^{-1}(-0.63)<0 \).
4. Hence \( C<A<B \).

Teacher's Note:
1. Decide only the sign and quadrant of each angle: negative, first quadrant, second quadrant.
2. cos inverse of a negative number always exceeds pi/2, and tan inverse of a negative number is negative.

Answer: (a) \( \mathbb{R} \)

Explanation:
1. \( \tan^{-1}t \) is defined for every real \( t \), and \( t=5x \) takes every real value as \( x \) varies over \( \mathbb{R} \).
2. So the domain is \( \mathbb{R} \); the interval \( \left(-\frac\pi2,\frac\pi2\right) \) is the range, not the domain.

Teacher's Note:
1. Do not confuse the domain with the range of tan inverse.
2. A linear change of argument does not restrict the domain of tan inverse.

Answer: (d) \( -\frac{\pi}{10} \)

Explanation:
1. \( \frac{33\pi}{5}=6\pi+\frac{3\pi}{5}\Rightarrow\cos\frac{33\pi}{5}=\cos\frac{3\pi}{5} \).
2. \( \cos\frac{3\pi}{5}=\sin\left(\frac\pi2-\frac{3\pi}{5}\right)=\sin\left(-\frac{\pi}{10}\right) \).
3. \( -\frac\pi{10}\in\left[-\frac\pi2,\frac\pi2\right] \), so the value is \( -\frac\pi{10} \).

Teacher's Note:
1. Remove full periods first, then convert cos to sin with the complementary angle.
2. The answer must lie in [-pi/2, pi/2], which rules out the first two options at once.

Answer: (a) \( \frac{4\pi}{7}+\frac{\pi}{55}-\frac{2\pi}{13} \)

Explanation:
1. \( \cos^{-1}\cos\frac{10\pi}{7}=\cos^{-1}\cos\left(2\pi-\frac{4\pi}{7}\right)=\frac{4\pi}{7} \).
2. \( \sin^{-1}\sin\frac{35\pi}{11}=\sin^{-1}\sin\left(\pi+\frac{2\pi}{11}\right)=-\frac{2\pi}{11} \).
3. \( \tan^{-1}\tan\frac{24\pi}{13}=\tan^{-1}\tan\left(\frac{24\pi}{13}-2\pi\right)=-\frac{2\pi}{13} \).
4. \( \cot^{-1}\cot\frac{26\pi}{5}=\cot^{-1}\cot\left(\frac{26\pi}{5}-5\pi\right)=\frac{\pi}{5} \).
5. Sum \( =\frac{4\pi}{7}-\frac{2\pi}{11}-\frac{2\pi}{13}+\frac{\pi}{5}=\frac{4\pi}{7}+\frac{\pi}{55}-\frac{2\pi}{13} \) since \( -\frac{2\pi}{11}+\frac{\pi}{5}=\frac{\pi}{55} \).

Teacher's Note:
1. Reduce each angle into the principal interval of its own inverse function, one term at a time (tan and cot have period pi, sin and cos have period 2 pi).
2. Combine the terms with denominators 11 and 5 first; they give pi/55.

Answer: (b) 2

Explanation:
1. \( \sin^{-1}\left(x^2-6x+8.5\right)=\frac\pi6\Rightarrow x^2-6x+8.5=\frac12 \).
2. \( x^2-6x+8=0\Rightarrow(x-2)(x-4)=0 \), so \( x=2 \) or \( x=4 \).
3. Among the options only \( x=2 \) appears.

Teacher's Note:
1. Isolate the inverse function, then take sine of both sides to get an ordinary quadratic.
2. Substituting the options is a good quick check: for x = 2 the bracket equals 4 - 12 + 8.5 = 0.5.

Answer: (d) \( \frac14 \)

Explanation:
1. \( \tan^{-1}(x+1)+\tan^{-1}(x-1)=\tan^{-1}\frac{2x}{2-x^2} \) (the product \( (x+1)(x-1)<1 \) for the valid root).
2. \( \frac{2x}{2-x^2}=\frac{8}{31}\Rightarrow62x=16-8x^2\Rightarrow4x^2+31x-8=0 \).
3. \( (4x-1)(x+8)=0\Rightarrow x=\frac14 \) (the root \( x=-8 \) gives \( xy>1 \) and is rejected).

Teacher's Note:
1. Check the condition xy < 1 before using the tan inverse addition formula without a pi correction.
2. With x = 1/4 the product (5/4)(-3/4) is negative, so the formula applies directly.

Answer: (c) \( \frac\pi4 \)

Explanation:
1. \( 2\tan^{-1}\frac56=\tan^{-1}\frac{5/3}{1-25/36}=\tan^{-1}\frac{60}{11} \).
2. \( \tan^{-1}\frac{60}{11}+\tan^{-1}\frac{11}{60}=\frac\pi2 \).
3. The given expression is half of \( 2\tan^{-1}\frac56+\tan^{-1}\frac{11}{60}=\frac\pi2 \), i.e. \( \frac\pi4 \).

Teacher's Note:
1. Factor out one half so that the first term becomes a double angle.
2. 60/11 and 11/60 are reciprocals, so their tan inverse values add up to pi/2.

Answer: (d) \( \frac{2b}{a} \)

Explanation:
1. Let \( \phi=\frac12\cos^{-1}\frac ab \), so \( \cos2\phi=\frac ab \).
2. \( \tan\left(\frac\pi4+\phi\right)+\tan\left(\frac\pi4-\phi\right)=\frac{1+\tan\phi}{1-\tan\phi}+\frac{1-\tan\phi}{1+\tan\phi}=\frac{2(1+\tan^2\phi)}{1-\tan^2\phi}=\frac{2}{\cos2\phi} \).
3. Hence the value is \( \frac{2b}{a} \).

Teacher's Note:
1. Combine the two fractions over a common denominator; the numerator becomes 2 (1 + tan squared) and the denominator 1 - tan squared.
2. The ratio (1 + tan squared)/(1 - tan squared) equals sec of the double angle.

Answer: (d) \( \frac{17\pi}{20} \)

Explanation:
1. \( \frac1{\sqrt2}\left(\cos\frac{9\pi}{10}-\sin\frac{9\pi}{10}\right)=\cos\frac{9\pi}{10}\cos\frac\pi4-\sin\frac{9\pi}{10}\sin\frac\pi4=\cos\left(\frac{9\pi}{10}+\frac\pi4\right)=\cos\frac{23\pi}{20} \).
2. \( \frac{23\pi}{20}\notin[0,\pi] \), so \( \cos\frac{23\pi}{20}=\cos\left(2\pi-\frac{23\pi}{20}\right)=\cos\frac{17\pi}{20} \).
3. Hence the value is \( \frac{17\pi}{20} \).

Teacher's Note:
1. Recognise the compound-angle pattern with cos 45 degrees and sin 45 degrees both equal to 1/root 2.
2. The combined angle 23 pi/20 is beyond pi, so reflect it with 2 pi minus the angle.

Answer: (c) \( \frac\pi4 \)

Explanation:
1. Let \( \theta=\cos^{-1}x\in\left(0,\frac\pi4\right) \). Then \( \frac{x+\sqrt{1-x^2}}{\sqrt2}=\cos\theta\cos\frac\pi4+\sin\theta\sin\frac\pi4=\cos\left(\theta-\frac\pi4\right) \).
2. Since \( \theta-\frac\pi4\in\left(-\frac\pi4,0\right) \), \( \cos^{-1}\cos\left(\theta-\frac\pi4\right)=\frac\pi4-\theta \).
3. The sum is \( \theta+\frac\pi4-\theta=\frac\pi4 \).

Teacher's Note:
1. The range 1/root 2 < x < 1 means theta lies between 0 and 45 degrees, so theta - 45 degrees is negative.
2. Taking absolute value of the angle inside cos inverse is what makes the theta terms cancel.

Answer: (d) \( \pi+\tan^{-1}\frac35 \)

Explanation:
1. \( \tan^{-1}2+\tan^{-1}3=\pi+\tan^{-1}\frac{5}{1-6}=\pi-\tan^{-1}1=\frac{3\pi}{4} \) (since the product exceeds 1).
2. \( \frac{3\pi}{4}+\tan^{-1}4=\pi+\left(\tan^{-1}4-\frac\pi4\right)=\pi+\tan^{-1}\frac{4-1}{1+4}=\pi+\tan^{-1}\frac35 \).

Teacher's Note:
1. For positive x and y with xy greater than 1, the sum of two tan inverse values exceeds pi/2, so a pi must be added.
2. Alternatively use the three-term formula: tangent of the sum is (sum - product)/(1 - sum of pair products) = 3/5.

Answer: (c) \( (\cot1,1] \)

Explanation:
1. Both greatest-integer terms are non-negative here, so each must be zero: \( 0\le\cot^{-1}x<1 \) and \( 0\le\cos^{-1}x<1 \).
2. Both functions are decreasing. \( \cot^{-1}x<1\Rightarrow x>\cot1\approx0.64 \) and \( \cos^{-1}x<1\Rightarrow x>\cos1\approx0.54 \).
3. The stronger condition is \( x>\cot1 \), and \( x\le1 \) for cos inverse to exist. So \( x\in(\cot1,1] \).

Teacher's Note:
1. A sum of two greatest-integer values equals 0 only when both values are 0 (for non-negative numbers).
2. Since cot 1 is larger than cos 1, the condition from cot inverse decides the lower end point.

Answer: (a) 0

Explanation:
1. Each of \( \cos^{-1}x \) and \( \cos^{-1}2x \) is non-negative, so the left side is at least \( \pi \).
2. It can never equal 0, so the equation has no real solution.

Teacher's Note:
1. Use the range [0, pi] of cos inverse to bound the left side before attempting any algebra.
2. Equations that are impossible by range arguments are common in competitive papers.

Answer: (b) \( \alpha<\gamma<\beta \)

Explanation:
1. \( \sqrt2-1=\tan22.5^\circ\Rightarrow\alpha=45^\circ=\frac\pi4 \).
2. \( \beta=3\cdot\frac\pi4-\frac\pi6=\frac{7\pi}{12}=105^\circ \).
3. Since \( \cos60^\circ=\frac12>\frac13>\cos90^\circ=0 \) and cosine decreases on \( [0^\circ,180^\circ] \), \( 60^\circ<\gamma<90^\circ \) (about \( 70.5^\circ \)).
4. Hence \( \alpha<\gamma<\beta \).

Teacher's Note:
1. Convert each quantity to degrees; it makes the comparison immediate.
2. gamma lies between 60 and 90 degrees because 1/3 lies between cos 90 degrees = 0 and cos 60 degrees = 1/2.

Answer: (a) \( 0 \)

Explanation:
1. \( \sin^2 15^\circ=\frac{1-\cos30^\circ}{2}=\frac{2-\sqrt3}{4}\Rightarrow\sin^{-1}\sqrt{\frac{2-\sqrt3}{4}}=\frac\pi{12} \).
2. \( \cos^{-1}\frac{\sqrt{12}}{4}=\cos^{-1}\frac{\sqrt3}{2}=\frac\pi6,\quad\sec^{-1}\sqrt2=\frac\pi4 \).
3. The sum is \( \frac{\pi}{12}+\frac{\pi}{6}+\frac{\pi}{4}=\frac\pi2 \), so \( \cot\frac\pi2=0 \) and \( \sin^{-1}0=0 \).

Teacher's Note:
1. Recognise (2 - root 3)/4 as sin squared of 15 degrees using the half-angle identity.
2. When the angle inside cot becomes pi/2 the cot vanishes, so the outer sin inverse gives 0.