JEE Mathematics Inverse Trigonometric Functions MCQs Set 16

Answer: (b) \(2\pi\)

Explanation:
1. For \(x>0\), \(\cot^{-1}(-x)=\pi-\cot^{-1}x\) and \(\cot^{-1}\dfrac{ab+1}{b-a}=\cot^{-1}a-\cot^{-1}b\) for \(b>a>0\), so the first term is \(\pi-(\cot^{-1}a-\cot^{-1}b)\) and the second is \(\pi-(\cot^{-1}b-\cot^{-1}c)\).
2. The third term is \(\cot^{-1}\dfrac{ca+1}{c-a}=\cot^{-1}a-\cot^{-1}c\). Adding: \(2\pi-(\cot^{-1}a-\cot^{-1}c)+(\cot^{-1}a-\cot^{-1}c)=2\pi\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (c) \(\left[-\dfrac12,\dfrac12\right]\)

Explanation:
1. With \(x=\sin\theta\), \(3x-4x^3=\sin3\theta\) and \(\sin^{-1}(\sin3\theta)=3\theta\) only when \(-\dfrac\pi2\le3\theta\le\dfrac\pi2\).
2. So \(-\dfrac\pi6\le\theta\le\dfrac\pi6\), i.e. \(-\dfrac12\le x\le\dfrac12\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (d) \(\dfrac{3\pi}4\)

Explanation:
1. \(\cot^{-1}2+\cot^{-1}3=\tan^{-1}\dfrac12+\tan^{-1}\dfrac13=\tan^{-1}\dfrac{5/6}{1-1/6}=\tan^{-1}1=\dfrac\pi4\).
2. Third angle \(=\pi-\dfrac\pi4=\dfrac{3\pi}4\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (a) \(\dfrac{x+y}{1-xy}\)

Explanation:
1. For \(|x|\le1\), \(\sin^{-1}\dfrac{2x}{1+x^2}=2\tan^{-1}x\); for \(y\ge0\), \(\cos^{-1}\dfrac{1-y^2}{1+y^2}=2\tan^{-1}y\).
2. The expression is \(\tan\left(\tan^{-1}x+\tan^{-1}y\right)=\dfrac{x+y}{1-xy}\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (c) \(5\)

Explanation:
1. \(\dfrac x\pi<\tan\dfrac\pi3=\sqrt3\), so \(x<\sqrt3\pi\approx5.44\).
2. The largest natural number is \(5\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (b) \(a=1\) and \(b=-\dfrac13\)

Explanation:
1. The first series sums to \(\dfrac{a}{1+a/3}=\dfrac{3a}{3+a}\) and the second to \(\dfrac1{1-b}\).
2. \(\sin^{-1}X+\cos^{-1}Y=\dfrac\pi2\) means \(X=Y\): \(\dfrac{3a}{3+a}=\dfrac1{1-b}\), which holds for \(a=1\), \(b=-\dfrac13\) (\(\dfrac34=\dfrac34\)).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (d) \(\dfrac\pi4\)

Explanation:
1. The \( n th \) term is \(\cot^{-1}(2n^2)=\tan^{-1}\dfrac1{2n^2}=\tan^{-1}\dfrac{2}{4n^2}=\tan^{-1}(2n+1)-\tan^{-1}(2n-1)\cdot\) in the form \(\tan^{-1}\dfrac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\).
2. The series telescopes to \(\tan^{-1}(\infty)-\tan^{-1}1=\dfrac\pi2-\dfrac\pi4=\dfrac\pi4\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (a) \(x=1,y=2\); \(x=2,y=7\)

Explanation:
1. \(\cos^{-1}\dfrac y{\sqrt{1+y^2}}=\tan^{-1}\dfrac1y\) and \(\sin^{-1}\dfrac3{\sqrt{10}}=\tan^{-1}3\), so \(\dfrac{x+1/y}{1-x/y}=3\), i.e. \(xy+1=3y-3x\).
2. \((x-3)(y+3)=-10\) gives \(y+3=5\) or \(10\) for positive \(y\), i.e. \((x,y)=(1,2)\) and \((2,7)\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (b) \((-\infty,\cot3]\)

Explanation:
1. The inequality is \(\left([\cot^{-1}x]-3\right)^2\le0\), so \([\cot^{-1}x]=3\), i.e. \(3\le\cot^{-1}x<\pi\) (as \(\cot^{-1}x<\pi<4\)).
2. \(\cot^{-1}\) is decreasing, so \(x\le\cot3\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (c) \(x\in\left[\dfrac{\sqrt2}2,1\right]\)

Explanation:
1. Let \(x=\cos\theta\), \(\theta\in[0,\pi]\); the right side is \(\sin^{-1}(\sin2\theta)\) and equals \(2\theta\) only if \(2\theta\in\left[-\dfrac\pi2,\dfrac\pi2\right]\).
2. So \(0\le\theta\le\dfrac\pi4\), giving \(\dfrac{\sqrt2}2\le x\le1\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (d) \(\dfrac16\)

Explanation:
1. \(\dfrac{5x}{1-6x^2}=1\) gives \(6x^2+5x-1=0\), so \(x=\dfrac16\) or \(x=-1\).
2. \(x=-1\) gives a negative left side (\(\tan^{-1}(-2)+\tan^{-1}(-3)<0\)), so only \(x=\dfrac16\) works.

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (a) \(2\)

Explanation:
1. \(y=\sqrt{1-x^2}\), so \(x^2+y^2=1\).
2. Numerator \(=1+(x^2+y^2)^2-2x^2y^2=2(1-x^2y^2)\) and denominator \(=1-x^2y^2\); the ratio is \(2\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (d) \(\dfrac1{\sqrt2}\)

Explanation:
1. \(\tan\dfrac{15\pi}4=\tan\left(4\pi-\dfrac\pi4\right)=-1\).
2. \(\tan^{-1}(-1)=-\dfrac\pi4\) and \(\cos\left(-\dfrac\pi4\right)=\dfrac1{\sqrt2}\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (c) \(4\sin^2\alpha\)

Explanation:
1. Let \(A=\cos^{-1}x\), \(B=\cos^{-1}\dfrac y2\), so \(x=\cos A\), \(y=2\cos B\) and \(A-B=\alpha\).
2. \(4\cos^2A-8\cos A\cos B\cos\alpha+4\cos^2B=4\left[\cos^2A+\cos^2B-2\cos A\cos B\cos(A-B)\right]=4\sin^2(A-B)\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (a) \(0\)

Explanation:
1. \(\sqrt{\dfrac{2-\sqrt3}4}=\sin15^\circ\), \(\dfrac{\sqrt{12}}4=\dfrac{\sqrt3}2\) and \(\sec^{-1}\sqrt2=\dfrac\pi4\).
2. The sum is \(\dfrac\pi{12}+\dfrac\pi6+\dfrac\pi4=\dfrac\pi2\), and \(\cot\dfrac\pi2=0\), so the value is \(\sin^{-1}0=0\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (b) \(\dfrac{146}{121}\)

Explanation:
1. \(\sec^2(\tan^{-1}t)=1+t^2\).
2. \(1+\dfrac{25}{121}=\dfrac{146}{121}\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (b) \(\dfrac\pi2\)

Explanation:
1. The \( n th \) term equals \(\tan^{-1}\sqrt n-\tan^{-1}\sqrt{n-1}\) (check \(n=1\): \(\dfrac\pi4-0\)).
2. The series telescopes to \(\lim_{n\to\infty}\tan^{-1}\sqrt n-\tan^{-1}0=\dfrac\pi2\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (c) \(1\)

Explanation:
1. The series sum to \(\dfrac x{1+x/2}\) and \(\dfrac{x^2}{1+x^2/2}\).
2. \(\sin^{-1}X+\cos^{-1}Y=\dfrac\pi2\) needs \(X=Y\): \(x\left(1+\dfrac{x^2}2\right)=x^2\left(1+\dfrac x2\right)\), giving \(x=1\) (as \(x\ne0\)).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (d) \(\cot^{-1}3\)

Explanation:
1. \(\cot^{-1}7+\cot^{-1}8=\cot^{-1}\dfrac{56-1}{15}=\cot^{-1}\dfrac{11}3\).
2. \(\cot^{-1}\dfrac{11}3+\cot^{-1}18=\cot^{-1}\dfrac{(11/3)(18)-1}{11/3+18}=\cot^{-1}\dfrac{65}{65/3}=\cot^{-1}3\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (a) \(\dfrac{a-b}{1+ab}\)

Explanation:
1. The left side is \(2\tan^{-1}a-2\tan^{-1}b\) and the right side is \(2\tan^{-1}x\).
2. So \(\tan^{-1}x=\tan^{-1}a-\tan^{-1}b\), giving \(x=\dfrac{a-b}{1+ab}\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (c) Two

Explanation:
1. \(\sqrt{1-x^2}=\dfrac12\) gives \(x=\pm\dfrac{\sqrt3}2\).
2. For \(x=\dfrac{\sqrt3}2\), \(\tan\dfrac\pi6=\dfrac1{\sqrt3}\); for \(x=-\dfrac{\sqrt3}2\), \(\tan\dfrac{5\pi}6=-\dfrac1{\sqrt3}\). So two values.

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (b) \(\dfrac\pi2\)

Explanation:
1. Let \(A=\tan^{-1}\dfrac{xy}{zr}\) etc. \(\tan(A+B)=\dfrac{\frac{xy}{zr}+\frac{yz}{xr}}{1-\frac{y^2}{r^2}}=\dfrac{y(x^2+z^2)}{xzr}\cdot\dfrac{r^2}{r^2-y^2}=\dfrac{yr}{xz}\).
2. Since \(\tan C=\dfrac{xz}{yr}\), \(\tan(A+B)=\cot C\), so \(A+B+C=\dfrac\pi2\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (c) \(2\)

Explanation:
1. With \(x=\sin A\), \(y=\sin B\), \(z=\sin C\) and \(A+B+C=\pi\) the known identity gives \(k=2\).
2. Check: \(x=y=z=\dfrac{\sqrt3}2\) gives left side \(\dfrac{54}{16}\) and \(\left(x^2y^2+y^2z^2+z^2x^2\right)=\dfrac{27}{16}\), so \(k=2\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (a) \(\dfrac{(n-1)d}{1+a_1a_n}\)

Explanation:
1. \(\tan^{-1}\dfrac d{1+a_ka_{k+1}}=\tan^{-1}a_{k+1}-\tan^{-1}a_k\).
2. The sum telescopes to \(\tan^{-1}a_n-\tan^{-1}a_1\), so the tangent is \(\dfrac{a_n-a_1}{1+a_1a_n}=\dfrac{(n-1)d}{1+a_1a_n}\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (b) \(\left[-\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right]\)

Explanation:
1. With \(x=\sin\theta\) the left side is \(\sin^{-1}(\sin2\theta)\), which equals \(2\theta\) only for \(2\theta\in\left[-\dfrac\pi2,\dfrac\pi2\right]\).
2. So \(|\theta|\le\dfrac\pi4\), i.e. \(|x|\le\dfrac1{\sqrt2}\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (c) \(k=\dfrac\pi4\), \(K=\dfrac{3\pi}4\)

Explanation:
1. \(x\in[-1,1]\) and \(\sin^{-1}x+\cos^{-1}x=\dfrac\pi2\), so the expression is \(\dfrac\pi2+\tan^{-1}x\).
2. \(\tan^{-1}x\) ranges over \(\left[-\dfrac\pi4,\dfrac\pi4\right]\), so the expression ranges over \(\left[\dfrac\pi4,\dfrac{3\pi}4\right]\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (d) \(\phi\)

Explanation:
1. \(\sin^{-1}(\sin4)=\pi-4<0\), because \(4\in\left(\dfrac\pi2,\dfrac{3\pi}2\right)\).
2. The quadratic needs \(k^2-4(\pi-4)<0\), i.e. \(k^2<4(\pi-4)<0\), which is impossible; the set is empty.

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (a) \(\dfrac\pi4\)

Explanation:
1. \(\tan(A+B)=\dfrac{2+3}{1-6}=-1\), and \(A+B\) lies in \(\left(\dfrac\pi2,\pi\right)\), so \(A+B=\dfrac{3\pi}4\).
2. \(C=\pi-\dfrac{3\pi}4=\dfrac\pi4\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (d) \(\dfrac15\)

Explanation:
1. \(\sin\theta=1\) gives \(\sin^{-1}\dfrac15+\cos^{-1}x=\dfrac\pi2\).
2. Since \(\sin^{-1}t+\cos^{-1}t=\dfrac\pi2\), \(x=\dfrac15\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.

Answer: (c) \(\dfrac\pi3\)

Explanation:
1. Let \(x=\cos\theta\) with \(0\le\theta\le\dfrac\pi3\); the bracket is \(\cos\theta\cos\dfrac\pi3+\sin\theta\sin\dfrac\pi3=\cos\left(\theta-\dfrac\pi3\right)\).
2. Its inverse cosine is \(\dfrac\pi3-\theta\), so the sum is \(\theta+\dfrac\pi3-\theta=\dfrac\pi3\).

Teacher's Note:
1. Use the principal value ranges: \( \sin^{-1} \) and \( \tan^{-1} \) lie in \( (-\pi/2, \pi/2) \) or \( [-\pi/2, \pi/2], \cos^{-1} \) and \( \cot^{-1} \) lie in \( [0, \pi] \) or \( (0, \pi) \).
2. Convert to a single inverse function with a substitution such as \( x = \tan(\theta) \) or \( x = \cos(\theta) \) and check the allowed interval.