JEE Mathematics Inverse Trigonometric Functions MCQs Set 01

Answer: (a) \( \frac{\sqrt{1-x^2}}{x} \)

Explanation:
1. \( \text{Let }\theta=\cos^{-1}x\Rightarrow \cos\theta=x,\ \sin\theta=\sqrt{1-x^2} \) (as \( 0\le\theta\le\pi \) the sine is non-negative).
2. \( \tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\sqrt{1-x^2}}{x} \).

Teacher's Note:
1. For inverse functions, name the angle and read the other ratios off a right triangle.
2. The result is valid for x not equal to 0, since cos 90 degrees makes the tangent undefined.

Answer: (b) \( \sqrt{\frac{x^2+1}{x^2+2}} \)

Explanation:
1. \( \cos(\tan^{-1}x)=\frac{1}{\sqrt{1+x^2}}=c \)
2. \( \sin(\cot^{-1}c)=\frac{1}{\sqrt{1+c^2}} \)
3. \( 1+c^2=\frac{x^2+2}{x^2+1}\Rightarrow \sin(\cdot)=\sqrt{\frac{x^2+1}{x^2+2}} \).

Teacher's Note:
1. Work from the inside out: convert each inverse function to an algebraic expression one layer at a time.
2. cot inverse of c is an angle in (0, pi), so its sine is positive and equals 1 over root (1 + c squared).

Answer: (c) \( 2\cot x\csc x \)

Explanation:
1. \( \tan 2\theta=\frac{2\tan\theta}{1-\tan^2\theta}\ \text{with}\ \tan\theta=\cos x \).
2. \( =\frac{2\cos x}{1-\cos^2x}=\frac{2\cos x}{\sin^2x}=2\cdot\frac{\cos x}{\sin x}\cdot\frac{1}{\sin x}=2\cot x\csc x \).

Teacher's Note:
1. Apply the double-angle formula for tan directly once the inner inverse is replaced by its angle.
2. Always simplify 1 minus cos squared to sin squared; it turns the denominator into a single trigonometric term.

Answer: (c) \( \frac{x^2+1}{x^2+2} \)

Explanation:
1. \( \sin(\cot^{-1}x)=\frac{1}{\sqrt{1+x^2}}=s \)
2. \( \cos^2(\tan^{-1}s)=\frac{1}{1+s^2} \)
3. \( =\frac{1}{1+\frac{1}{1+x^2}}=\frac{x^2+1}{x^2+2} \).

Teacher's Note:
1. cos squared of tan inverse s equals 1 over (1 + s squared); remember this as a ready formula.
2. The answer lies strictly between 0 and 1, which is a quick sanity check against the other options.

Answer: (c) \( \pi \)

Explanation:
1. \( \cot^{-1}\frac{ab+1}{a-b}=\tan^{-1}\frac{a-b}{1+ab}=\tan^{-1}a-\tan^{-1}b \) (positive argument).
2. Similarly the second term is \( \tan^{-1}b-\tan^{-1}c \).
3. In the third term \( \frac{ca+1}{c-a}<0 \), so \( \cot^{-1}(\cdot)=\pi+\tan^{-1}\frac{c-a}{1+ca}=\pi+\tan^{-1}c-\tan^{-1}a \).
4. Adding, everything cancels except \( \pi \).

Teacher's Note:
1. cot inverse of a negative number lies in (pi/2, pi), so a pi must be added when converting to tan inverse.
2. Telescoping sums of tan inverse differences are a common JEE pattern.

Answer: (a) \( 0 \)

Explanation:
1. \( 2\cos^2\theta-1=1-2\sin^2\theta=\cos2\theta=y \).
2. \( \sin^{-1}y+\cos^{-1}y=\frac{\pi}{2}\ \text{for}\ |y|\le1 \).
3. \( \cos\frac{\pi}{2}=0 \).

Teacher's Note:
1. Spot that both arguments are the same quantity cos 2 theta.
2. sin inverse y + cos inverse y = pi/2 is the single most useful identity in this chapter.

Answer: (a) \( xy+yz+zx=1 \)

Explanation:
1. \( \tan^{-1}x+\tan^{-1}y=\frac{\pi}{2}-\tan^{-1}z=\cot^{-1}z=\tan^{-1}\frac{1}{z} \).
2. \( \frac{x+y}{1-xy}=\frac{1}{z}\Rightarrow z(x+y)=1-xy\Rightarrow xy+yz+zx=1 \).

Teacher's Note:
1. Sum equal to pi/2 gives xy + yz + zx = 1, while sum equal to pi gives x + y + z = xyz.
2. Compare with the tan A + tan B + tan C = tan A tan B tan C result for a triangle.

Answer: (a) \( \sin^2\theta \)

Explanation:
1. \( \text{Let }A=\cos^{-1}\frac{x}{a},\ B=\cos^{-1}\frac{y}{b} \), so \( A+B=\theta \).
2. \( \cos\theta=\cos A\cos B-\sin A\sin B=\frac{xy}{ab}-\sqrt{1-\frac{x^2}{a^2}}\sqrt{1-\frac{y^2}{b^2}} \).
3. Move the product term across and square: \( \left(\frac{xy}{ab}-\cos\theta\right)^2=\left(1-\frac{x^2}{a^2}\right)\left(1-\frac{y^2}{b^2}\right) \).
4. Expanding and cancelling gives \( \frac{x^2}{a^2}-\frac{2xy}{ab}\cos\theta+\frac{y^2}{b^2}=1-\cos^2\theta=\sin^2\theta \).

Teacher's Note:
1. This is the standard elimination result for the sum of two cos inverse terms.
2. When a = b = 1 it reduces to x squared - 2xy cos theta + y squared = sin squared theta.

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

Explanation:
1. Put \( x=\sin\theta \) so \( 2x\sqrt{1-x^2}=\sin2\theta \).
2. \( \sin^{-1}(\sin2\theta)=2\theta \) only if \( -\frac{\pi}{2}\le2\theta\le\frac{\pi}{2} \), i.e. \( -\frac{\pi}{4}\le\theta\le\frac{\pi}{4} \).
3. Hence \( -\frac{1}{\sqrt2}\le x\le\frac{1}{\sqrt2} \).

Teacher's Note:
1. The range of sin inverse is [-pi/2, pi/2], so the doubled angle must stay in that interval.
2. Outside this interval the identity changes to pi - 2 sin inverse x or -pi - 2 sin inverse x.

Answer: (d) \( x=n\pi,\ n\in\mathbb{Z} \)

Explanation:
1. The domain of \( \operatorname{cosec}^{-1} \) is \( |t|\ge1 \).
2. Since \( |\cos x|\le1 \), we need \( |\cos x|=1 \).
3. \( \cos x=\pm1\Rightarrow x=n\pi,\ n\in\mathbb{Z} \).

Teacher's Note:
1. The domain of cosec inverse and sec inverse is |t| >= 1, the complement of the open interval (-1, 1).
2. A function bounded by 1 in modulus can enter that domain only at the boundary values +1 and -1.

Answer: (d) All of the above

Explanation:
1. Put \( x=\cos\theta,\ 0<\theta<\frac{\pi}{2} \). Then \( \frac{\sqrt{1-x^2}}{1+x}=\frac{\sin\theta}{1+\cos\theta}=\tan\frac{\theta}{2} \).
2. So the expression equals \( \frac{\theta}{2}=\frac12\cos^{-1}x \).
3. \( \cos\frac{\theta}{2}=\sqrt{\frac{1+x}{2}},\ \sin\frac{\theta}{2}=\sqrt{\frac{1-x}{2}} \), so options (b) and (c) are the same angle too.

Teacher's Note:
1. Half-angle substitution x = cos theta removes the radical neatly.
2. When every listed option is an equivalent form of the same angle, the answer is All of the above.

Answer: (a) \( (0,1] \)

Explanation:
1. Need \( \sin^{-1}x \) defined: \( -1\le x\le1 \).
2. Need \( \sin^{-1}x>0\Rightarrow x>0 \).
3. Combining: \( 0\lt x\le1 \).

Teacher's Note:
1. A logarithm needs a strictly positive argument, so the inverse sine must be positive.
2. Always intersect the domain of the inner function with the condition imposed by the outer function.

Answer: (c) \( \frac{5\pi}{6} \)

Explanation:
1. \( \cos\frac{7\pi}{6}=\cos\left(2\pi-\frac{5\pi}{6}\right)=\cos\frac{5\pi}{6} \).
2. \( \frac{5\pi}{6}\in[0,\pi] \), so \( \cos^{-1}\left(\cos\frac{7\pi}{6}\right)=\frac{5\pi}{6} \).

Teacher's Note:
1. cos inverse of cos theta equals theta only when theta is in [0, pi].
2. Otherwise reduce theta to an equivalent angle in [0, pi] having the same cosine.

Answer: (b) \( \frac{2\pi}{3} \)

Explanation:
1. \( \sin^{-1}\frac{\sqrt3}{2}=\frac{\pi}{3},\ \tan^{-1}(-\sqrt3)=-\frac{\pi}{3} \).
2. \( \frac{\pi}{3}-\left(-\frac{\pi}{3}\right)=\frac{2\pi}{3} \).

Teacher's Note:
1. tan inverse is an odd function, so tan inverse of a negative number is the negative of the principal value.
2. Learn the principal values for 0, 1/2, 1/root 2, root 3/2 and 1 to answer such questions at sight.

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

Explanation:
1. \( \cot\frac{\pi}{12}=\cot15^\circ=2+\sqrt3 \).
2. Since \( \frac{\pi}{12}\in(0,\pi) \), \( \cot^{-1}(2+\sqrt3)=\frac{\pi}{12} \).

Teacher's Note:
1. cot 15 degrees = 2 + root 3 and tan 15 degrees = 2 - root 3 are worth memorising.
2. The principal range of cot inverse is (0, pi).

Answer: (c) 15

Explanation:
1. \( \cot\alpha=\frac12\Rightarrow\tan\alpha=2\Rightarrow\sec^2\alpha=1+4=5 \).
2. \( \tan\beta=\frac13\Rightarrow\cot\beta=3\Rightarrow\operatorname{cosec}^2\beta=1+9=10 \).
3. Sum \( =15 \).

Teacher's Note:
1. sec squared = 1 + tan squared and cosec squared = 1 + cot squared are all that is needed here.
2. Convert each inverse function to its reciprocal ratio carefully before squaring.

Answer: (b) \( \frac{117}{125} \)

Explanation:
1. \( \cos A=\frac35,\ \cos B=\frac{24}{25} \) (both angles acute).
2. \( \sin(A+B)=\frac45\cdot\frac{24}{25}+\frac35\cdot\frac{7}{25}=\frac{96+21}{125}=\frac{117}{125} \).

Teacher's Note:
1. Use sin(A + B) = sin A cos B + cos A sin B after finding the missing cosines from a right triangle.
2. Both angles are in the first quadrant, so the cosines are positive.

Answer: (b) \( -\frac{527}{625} \)

Explanation:
1. \( \cos2\theta=2\cos^2\theta-1=2\cdot\frac{49}{625}-1=-\frac{527}{625} \).

Teacher's Note:
1. With cos theta given, use cos 2 theta = 2 cos squared theta - 1 and no need to find theta.
2. A negative value shows that 2 theta lies in the second quadrant, which is consistent with theta being greater than 45 degrees.

Answer: (b) \( \frac{\pi}{2} \)

Explanation:
1. \( x+\frac1x=2\Rightarrow(x-1)^2=0\Rightarrow x=1 \).
2. \( \sin^{-1}1=\frac{\pi}{2} \).

Teacher's Note:
1. x + 1/x equals 2 only for x = 1 when x is positive, by AM-GM equality.
2. The principal value of sin inverse always lies in [-pi/2, pi/2], so options (c) and (d) can be discarded at once.

Answer: (d) \( -\frac{7}{17} \)

Explanation:
1. \( \tan\left(2\tan^{-1}\tfrac15\right)=\frac{2/5}{1-1/25}=\frac{5}{12} \).
2. \( \tan\left(A-\frac{\pi}{4}\right)=\frac{\frac5{12}-1}{1+\frac5{12}}=\frac{-7/12}{17/12}=-\frac{7}{17} \).

Teacher's Note:
1. First find tan of the doubled angle, then use the tan(A - B) formula with tan 45 degrees = 1.
2. The sign check: 2 tan inverse (1/5) is about 22.6 degrees, which is less than 45 degrees, so the tangent of the difference is negative.

Answer: (c) \( \frac{\pi}{2} \)

Explanation:
1. \( 2\tan^{-1}\frac12=\tan^{-1}\frac{2\cdot\frac12}{1-\frac14}=\tan^{-1}\frac43 \).
2. \( \sin^{-1}\frac35=\tan^{-1}\frac34 \).
3. \( \tan^{-1}\frac43+\tan^{-1}\frac34=\frac{\pi}{2} \) because the two arguments are reciprocals of each other and positive.

Teacher's Note:
1. tan inverse x + tan inverse (1/x) = pi/2 for x > 0.
2. Converting sin inverse to tan inverse using a 3-4-5 triangle makes the pairing visible.

Answer: (c) \( \frac{3\pi}{4} \)

Explanation:
1. Here \( xy=6>1 \) and both are positive, so \( \tan^{-1}x+\tan^{-1}y=\pi+\tan^{-1}\frac{x+y}{1-xy} \).
2. \( =\pi+\tan^{-1}\frac{5}{-5}=\pi+\tan^{-1}(-1)=\pi-\frac{\pi}{4}=\frac{3\pi}{4} \).

Teacher's Note:
1. When x, y > 0 and xy > 1 the sum of the two tan inverse terms exceeds pi/2, so pi is added.
2. Check: tan inverse 2 is about 63.4 degrees and tan inverse 3 is about 71.6 degrees, sum 135 degrees.

Answer: (c) \( \frac{\pi}{4} \)

Explanation:
1. \( \tan(A-B)=\frac{\frac mn-\frac{m-n}{m+n}}{1+\frac mn\cdot\frac{m-n}{m+n}}=\frac{m(m+n)-n(m-n)}{n(m+n)+m(m-n)}=\frac{m^2+n^2}{m^2+n^2}=1 \).
2. For positive m, n the difference lies in \( (0,\pi/2) \), hence it equals \( \frac{\pi}{4} \).

Teacher's Note:
1. Standard result: the difference of these two angles is pi/4.
2. The arguments are the tangents of theta and theta - 45 degrees, which is why the answer is constant.

Answer: (c) \( \frac{\pi}{4} \)

Explanation:
1. \( 2\tan^{-1}\frac13=\tan^{-1}\frac{2/3}{1-1/9}=\tan^{-1}\frac34 \).
2. \( \tan^{-1}\frac34+\tan^{-1}\frac17=\tan^{-1}\frac{\frac34+\frac17}{1-\frac{3}{28}}=\tan^{-1}\frac{25/28}{25/28}=\tan^{-1}1=\frac{\pi}{4} \).

Teacher's Note:
1. Combine step by step, checking that the product of the arguments stays below 1 so that no pi has to be added.
2. Here 3/4 times 1/7 is less than 1, so the simple formula applies.

Answer: (c) \( 3\pi-10 \)

Explanation:
1. \( \sin10=\sin(\pi-10)=\sin(3\pi-10) \) because \( \sin(3\pi-\theta)=\sin\theta \).
2. \( 3\pi-10\approx-0.575\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right] \), so \( \sin^{-1}(\sin10)=3\pi-10 \).

Teacher's Note:
1. Shift 10 radians by a multiple of pi so that the angle falls in [-pi/2, pi/2].
2. 3 pi is about 9.42, so 10 lies just beyond 3 pi; the equivalent angle is the small negative number 3 pi - 10.