JEE Mathematics Inverse Trigonometric Functions MCQs Set 04

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

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

Teacher's Note:
1. Use the principal values and the identities cos⁻¹(-x) = π - cos⁻¹ x and tan⁻¹(-x) = -tan⁻¹ x. Bring all terms to a common denominator.

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

Explanation:
1. \( \sec^{-1}\frac1x=\cos^{-1}x \), so the left side is \( \frac{\sqrt{1-x^2}}{x} \).
2. \( \sin(\tan^{-1}2)=\frac2{\sqrt5} \).
3. \( 5(1-x^2)=4x^2 \), so \( x^2=\frac59 \) and \( x=\frac{\sqrt5}3 \).

Teacher's Note:
1. Convert each inverse function to a right triangle. Square both sides carefully and keep the positive root since x > 0.

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

Explanation:
1. Write \( \frac x2+\frac{\sqrt3}2\sqrt{1-x^2}=x\cos\frac\pi3+\sqrt{1-x^2}\sin\frac\pi3=\cos\left(\cos^{-1}x-\frac\pi3\right) \).
2. For \( \frac12\le x\le1 \), \( \cos^{-1}x\le\frac\pi3 \), so the inverse cosine gives \( \frac\pi3-\cos^{-1}x \).
3. Adding \( \cos^{-1}x \) gives \( \frac\pi3 \).

Teacher's Note:
1. The range restriction decides the sign inside the principal value. Check it at x = 1/2 and x = 1.

Answer: (b) \( \frac6{17} \)

Explanation:
1. \( \operatorname{cosec}^{-1}\frac53=\sin^{-1}\frac35=\tan^{-1}\frac34 \).
2. \( \tan^{-1}\frac34+\tan^{-1}\frac23=\tan^{-1}\frac{\frac34+\frac23}{1-\frac12}=\tan^{-1}\frac{17}6 \).
3. \( \cot\left(\tan^{-1}\frac{17}6\right)=\frac6{17} \).

Teacher's Note:
1. Change cosec⁻¹ to tan⁻¹ via a right triangle with sides 3, 4, 5. The product of the tangents is 1/2 which is less than 1, so the formula applies directly.

Answer: (d) \( \frac\pi2 \)

Explanation:
1. \( 2\tan^{-1}\frac13=\tan^{-1}\frac{2/3}{1-1/9}=\tan^{-1}\frac34 \).
2. \( \tan^{-1}\frac34=\cos^{-1}\frac45 \).
3. \( \sin^{-1}\frac45+\cos^{-1}\frac45=\frac\pi2 \).

Teacher's Note:
1. Use the double-angle formula for tan⁻¹ when the argument has modulus less than 1. Then convert to cos⁻¹ to apply sin⁻¹ x + cos⁻¹ x = π/2.

Answer: (b) \( \left(\frac1{\sqrt2},1\right] \)

Explanation:
1. \( \cos^{-1}x=\frac\pi2-\sin^{-1}x \).
2. The inequality becomes \( 2\sin^{-1}x>\frac\pi2 \), i.e. \( \sin^{-1}x>\frac\pi4 \).
3. Since sine increases on [-1, 1] in the principal range, \( x>\frac1{\sqrt2} \), and \( x\le1 \).

Teacher's Note:
1. Replace cos⁻¹ by π/2 minus sin⁻¹ so that only one inverse function remains. The domain restriction x ≤ 1 gives the upper end.

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

Explanation:
1. \( \cot^2y=\operatorname{cosec}^2y-1=(r^2+1)(r^2+2r+2)-1=(r^2+r+1)^2 \).
2. \( \tan y=\frac1{r^2+r+1}=\frac{(r+1)-r}{1+r(r+1)} \), so \( y=\tan^{-1}(r+1)-\tan^{-1}r \).
3. The sum telescopes: \( \tan^{-1}(n+1)-\tan^{-1}1=\tan^{-1}(n+1)-\frac\pi4 \).

Teacher's Note:
1. Use cosec² - cot² = 1 to convert to a tangent of a difference. The series then telescopes. Check n = 1: tan⁻¹ 2 - π/4 = tan⁻¹(1/3).

Answer: (c) \( 5-2\pi \)

Explanation:
1. 5 radians lies outside \( [-\frac\pi2,\frac\pi2] \).
2. \( \sin(5-2\pi)=\sin5 \) and \( 5-2\pi\approx-1.28 \) lies inside the principal range.
3. So \( \sin^{-1}(\sin5)=5-2\pi \).

Teacher's Note:
1. Shift the angle by a multiple of 2 π (or reflect) until it lies in [-π/2, π/2]. The result is negative here because 5 is in the fourth quadrant.

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

Explanation:
1. The principal range of cos⁻¹ is [0, π].
2. \( \cos(4\pi-10)=\cos10 \) and \( 4\pi-10\approx2.57 \) lies in [0, π].
3. So \( \cos^{-1}(\cos10)=4\pi-10 \).

Teacher's Note:
1. Find the angle in [0, π] with the same cosine by using the periodicity and the evenness of cosine.

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

Explanation:
1. \( \tan(-6)=\tan(2\pi-6) \) because tangent has period \( \pi \).
2. \( 2\pi-6\approx0.283 \) lies in \( (-\frac\pi2,\frac\pi2) \).
3. So the value is \( 2\pi-6 \).

Teacher's Note:
1. Add a multiple of π to bring the angle inside the principal range of tan⁻¹. Here adding 2 π works.

Answer: (c) \( x + y + z = xyz \)

Explanation:
1. Taking tangent of \( \tan^{-1}x+\tan^{-1}y=\pi-\tan^{-1}z \): \( \frac{x+y}{1-xy}=-z \).
2. So \( x+y=-z+xyz \), i.e. \( x+y+z=xyz \).

Teacher's Note:
1. Move one term to the right side and take the tangent. The fact tan(π - A) = -tan A is used.

Answer: (d) \( \sqrt{ab} \)

Explanation:
1. The sum of two positive inverse tangents is \( \frac\pi2 \) when the product of the arguments is 1.
2. \( \frac ax\cdot\frac bx=1 \), so \( x^2=ab \).

Teacher's Note:
1. Since tan⁻¹ u + tan⁻¹ v = π/2 means uv = 1 for positive u, v. Take the positive root.

Answer: (c) \( \sqrt{a^2+b^2} \)

Explanation:
1. \( \sin^{-1}\frac bx=\frac\pi2-\sin^{-1}\frac ax=\cos^{-1}\frac ax \).
2. So \( \frac bx=\sqrt{1-\frac{a^2}{x^2}} \), giving \( b^2=x^2-a^2 \).
3. \( x=\sqrt{a^2+b^2} \).

Teacher's Note:
1. Convert the second term to a cos⁻¹ of the first argument. The relation is a right triangle with hypotenuse x.

Answer: (b) \( xy = 1 \)

Explanation:
1. \( \tan^{-1}y=\frac\pi2-\tan^{-1}x=\cot^{-1}x=\tan^{-1}\frac1x \) for x > 0.
2. So \( y=\frac1x \), i.e. xy = 1.

Teacher's Note:
1. The condition is symmetric in x and y. For positive numbers it describes reciprocals.

Answer: (b) 3

Explanation:
1. Each \( \cos^{-1} \) is at most \( \pi \), so the sum 3 π forces each to equal π.
2. Hence x = y = z = -1.
3. \( xy+yz+zx=1+1+1=3 \).

Teacher's Note:
1. The maximum value of each cos⁻¹ term is π, so equality in the total forces every term to its maximum.

Answer: (a) \( \frac{\pi ab+c(a-b)}{a+b} \)

Explanation:
1. Let \( s=\sin^{-1}x \) so that \( \cos^{-1}x=\frac\pi2-s \).
2. From the given relation: \( (a+b)s=c+\frac{b\pi}2 \).
3. Then \( as+b(\frac\pi2-s)=(a-b)s+\frac{b\pi}2=\frac{(a-b)(c+b\pi/2)}{a+b}+\frac{b\pi}2 \).
4. Simplifying gives \( \frac{\pi ab+c(a-b)}{a+b} \).

Teacher's Note:
1. Express both inverse functions through one variable s. Check with a = b: both sides reduce to b π/2.

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

Explanation:
1. Let \( \alpha=\cos^{-1}\frac xa \), \( \beta=\cos^{-1}\frac yb \), so \( \cos\theta=\cos(\alpha+\beta)=\frac{xy}{ab}-\sqrt{1-\frac{x^2}{a^2}}\sqrt{1-\frac{y^2}{b^2}} \).
2. Move the product term and square: \( (\frac{xy}{ab}-\cos\theta)^2=(1-\frac{x^2}{a^2})(1-\frac{y^2}{b^2}) \).
3. Expanding gives \( \frac{x^2}{a^2}-\frac{2xy}{ab}\cos\theta+\frac{y^2}{b^2}=\sin^2\theta \).

Teacher's Note:
1. Take the cosine of both sides, isolate the square-root product and square. The terms x² y²/(a² b²) cancel.

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

Explanation:
1. Let \( \alpha=\sin^{-1}\frac xa \), \( \beta=\sin^{-1}\frac yb \). \( \cos\theta=\cos\alpha\cos\beta-\sin\alpha\sin\beta \).
2. So \( \cos\theta+\frac{xy}{ab}=\sqrt{1-\frac{x^2}{a^2}}\sqrt{1-\frac{y^2}{b^2}} \).
3. Squaring and simplifying gives \( \frac{x^2}{a^2}+\frac{2xy}{ab}\cos\theta+\frac{y^2}{b^2}=\sin^2\theta \).

Teacher's Note:
1. The sign of the cross term is positive here, in contrast with the cos⁻¹ version. Test with α = β = π/4, a = b = 1.

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

Explanation:
1. \( \tan^{-1}\frac1{1+r(r+1)}=\tan^{-1}\frac{(r+1)-r}{1+r(r+1)}=\tan^{-1}(r+1)-\tan^{-1}r \).
2. Summing from r = 1 to n telescopes to \( \tan^{-1}(n+1)-\tan^{-1}1 \).
3. \( =\tan^{-1}(n+1)-\frac\pi4 \).

Teacher's Note:
1. Write 1 as (r + 1) - r in the numerator to see the difference of tangents. The terms then cancel in pairs.

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

Explanation:
1. Let \( \theta=\cos^{-1}x \), so \( \cos\theta=x \) and \( \sin\theta=\sqrt{1-x^2} \).
2. \( \tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\sqrt{1-x^2}}x \).

Teacher's Note:
1. Draw a right triangle with adjacent side x and hypotenuse 1. The opposite side is √(1 - x²).

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

Explanation:
1. \( \cos(\tan^{-1}x)=\frac1{\sqrt{1+x^2}}=c \).
2. \( \sin(\cot^{-1}c)=\frac1{\sqrt{1+c^2}} \).
3. \( =\frac1{\sqrt{1+\frac1{1+x^2}}}=\sqrt{\frac{x^2+1}{x^2+2}} \).

Teacher's Note:
1. Work from the inside outwards, converting each inverse function with a right triangle.

Answer: (b) \( 2\cot x\operatorname{cosec}x \)

Explanation:
1. \( \tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta} \) with \( \tan\theta=\cos x \).
2. \( =\frac{2\cos x}{1-\cos^2x}=\frac{2\cos x}{\sin^2x} \).
3. \( =2\cot x\operatorname{cosec}x \).

Teacher's Note:
1. Apply the double-angle formula and use 1 - cos² x = sin² x. Then split into cot x and cosec x.

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

Explanation:
1. \( \sin(\cot^{-1}x)=\frac1{\sqrt{1+x^2}}=s \).
2. \( \cos^2(\tan^{-1}s)=\frac1{1+s^2}=\frac1{1+\frac1{1+x^2}} \).
3. \( =\frac{x^2+1}{x^2+2} \).

Teacher's Note:
1. Use cos²(tan⁻¹ s) = 1/(1 + s²). Simplify the compound fraction carefully.

Answer: (b) \( \pi \)

Explanation:
1. \( \cot^{-1}\frac{ab+1}{a-b}=\tan^{-1}a-\tan^{-1}b \) (a > b), and similarly \( \tan^{-1}b-\tan^{-1}c \).
2. Since c - a is negative, \( \cot^{-1}\frac{ca+1}{c-a}=\pi-\cot^{-1}\frac{ca+1}{a-c}=\pi-(\tan^{-1}a-\tan^{-1}c) \).
3. Adding the three terms gives \( \pi \).

Teacher's Note:
1. The third term has a negative argument, so its cot⁻¹ lies between π/2 and π. Use cot⁻¹(-t) = π - cot⁻¹ t.

Answer: (c) \( x + y + z = xyz \)

Explanation:
1. For positive x: \( \sec^{-1}\sqrt{1+x^2}=\tan^{-1}x \), \( \operatorname{cosec}^{-1}\frac{\sqrt{1+y^2}}y=\tan^{-1}y \), \( \cot^{-1}\frac1z=\tan^{-1}z \).
2. So \( \tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\pi \).
3. This gives \( x+y+z=xyz \).

Teacher's Note:
1. Use a right triangle for each term. Then apply the three-angle identity with the sum π.

Answer: (c) \( \frac{x-y}{1+xy} \)

Explanation:
1. \( \sin^{-1}\frac{2x}{1+x^2}=2\tan^{-1}x \) for \( |x|\le1 \), and \( \cos^{-1}\frac{1-y^2}{1+y^2}=2\tan^{-1}y \) for \( y\ge0 \).
2. The bracket becomes \( \tan(\tan^{-1}x-\tan^{-1}y) \).
3. \( =\frac{x-y}{1+xy} \).

Teacher's Note:
1. The half angles remove the factor 2. Check the domain conditions before using the standard conversions.

Answer: (d) 0

Explanation:
1. \( 2\cos^2\theta-1=1-2\sin^2\theta=\cos2\theta \).
2. The bracket is \( \sin^{-1}u+\cos^{-1}u=\frac\pi2 \) with \( u=\cos2\theta \).
3. \( \cos\frac\pi2=0 \).

Teacher's Note:
1. Recognise both arguments as cos 2 θ. The identity sin⁻¹ u + cos⁻¹ u = π/2 then gives the result.