JEE Mathematics Inverse Trigonometric Functions MCQs Set 06

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

Explanation:
1. \( \frac{2\sqrt2}{1+\sqrt3}=\sqrt2(\sqrt3-1)=\sqrt6-\sqrt2 \).
2. \( \cos\theta=\frac1{\sqrt6-\sqrt2}=\frac{\sqrt6+\sqrt2}4=\cos15^\circ \).
3. So \( \theta=\frac\pi{12} \).

Teacher's Note:
1. Rationalise the denominator and recognise cos 15 degrees = (√6 + √2)/4.

Answer: (b) \( \frac{14}{15} \)

Explanation:
1. \( \sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{2/3}{10/9}=\frac35 \).
2. \( \cos(\tan^{-1}2\sqrt2)=\frac1{\sqrt{1+8}}=\frac13 \).
3. Sum = \( \frac35+\frac13=\frac{14}{15} \).

Teacher's Note:
1. Use the formula sin 2 θ in terms of tan θ. Then evaluate the cosine with a right triangle.

Answer: (c) 10

Explanation:
1. \( \sec^2(\tan^{-1}2)=1+4=5 \).
2. \( \operatorname{cosec}^2(\cot^{-1}2)=1+\cot^2(\cot^{-1}2)=1+4=5 \).
3. Total = 10.

Teacher's Note:
1. Each inverse function cancels with the matching trigonometric function inside the identity.

Answer: (a) \( \cos^{-1}\frac{204}{325} \)

Explanation:
1. \( \cos(A+B)=\frac45\cdot\frac{63}{65}-\frac35\cdot\frac{16}{65}=\frac{252-48}{325}=\frac{204}{325} \).
2. A + B is less than π (about 0.89 rad), so the sum equals \( \cos^{-1}\frac{204}{325} \).

Teacher's Note:
1. Use cos(A + B) = cos A cos B - sin A sin B. Check that the sum lies in [0, π] before inverting.

Answer: (d) \( \frac2{\sqrt5} \)

Explanation:
1. Let \( \phi=\cot^{-1}(-\frac34)\in(\frac\pi2,\pi) \), so \( \cos\phi=-\frac35 \) and \( \sin\phi=\frac45 \).
2. \( \sin\frac\phi2=\sqrt{\frac{1-\cos\phi}2}=\sqrt{\frac{8/5}2}=\frac2{\sqrt5} \).
3. The half angle lies in the first quadrant, so the sine is positive.

Teacher's Note:
1. A negative argument puts cot⁻¹ in the second quadrant. Use the half-angle formula with the correct sign.

Answer: (a) an irrational number

Explanation:
1. \( \cos^{-1}x \) needs \( |x|\le1 \) and \( \sec^{-1}x \) needs \( |x|\ge1 \), so \( x=-1 \).
2. \( \cos^{-1}(-1)=\pi \) and \( \sec^{-1}(-1)=\pi \).
3. The sum is \( 2\pi \), an irrational number.

Teacher's Note:
1. Intersect the two domains to find the only admissible x. Then evaluate.

Answer: (d) 2

Explanation:
1. Let \( t=\frac{\sqrt5-1}2 \); then \( t^2+t-1=0 \), so \( 1-t^2=t \).
2. \( \tan2\theta=\frac{2t}{1-t^2}=\frac{2t}t=2 \).

Teacher's Note:
1. The number t is the golden-ratio conjugate, which satisfies t² = 1 - t. This simplifies the fraction.

Answer: (b) \( \frac\pi4 \)

Explanation:
1. The product \( \frac12\cdot\frac13=\frac16<1 \).
2. \( \tan^{-1}\frac{\frac12+\frac13}{1-\frac16}=\tan^{-1}1=\frac\pi4 \).

Teacher's Note:
1. The product of the arguments is less than 1, so no π term is added.

Answer: (b) \( \frac\pi4 \)

Explanation:
1. \( \tan^{-1}\frac{\frac32-\frac15}{1+\frac3{10}}=\tan^{-1}\frac{13/10}{13/10}=\tan^{-1}1 \).
2. \( =\frac\pi4 \).

Teacher's Note:
1. The product 3/10 is positive, so the difference formula applies without a π term.

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

Explanation:
1. \( 4\tan^{-1}\frac15=\tan^{-1}\frac{120}{119} \).
2. \( \tan^{-1}\frac{120}{119}-\tan^{-1}\frac1{239}=\tan^{-1}\frac{\frac{120}{119}-\frac1{239}}{1+\frac{120}{119\cdot239}}=\tan^{-1}1 \).
3. The value is \( \frac\pi4 \).

Teacher's Note:
1. This is the classical Machin formula for π. Compute 4 tan⁻¹(1/5) in two steps.

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

Explanation:
1. \( 2\tan^{-1}\frac15=\tan^{-1}\frac5{12} \) and \( 2\tan^{-1}\frac18=\tan^{-1}\frac{16}{63} \).
2. \( \tan^{-1}\frac5{12}+\tan^{-1}\frac{16}{63}=\tan^{-1}\frac{507/756}{676/756}=\tan^{-1}\frac34 \).
3. So the sum is \( \tan^{-1}\frac34+\tan^{-1}\frac17=\tan^{-1}\frac{\frac34+\frac17}{1-\frac3{28}}=\tan^{-1}1=\frac\pi4 \).

Teacher's Note:
1. Pair the terms so that every partial sum has product below 1. Check numerically: 0.3948 + 0.1419 + 0.2487 = 0.7854.

Answer: (d) \( \frac{\sqrt3}2 \)

Explanation:
1. \( \cos^{-1}x=\frac\pi2-\sin^{-1}x \), so \( 2\sin^{-1}x-\frac\pi2=\frac\pi6 \).
2. \( \sin^{-1}x=\frac\pi3 \).
3. \( x=\sin\frac\pi3=\frac{\sqrt3}2 \).

Teacher's Note:
1. Express the equation through one inverse function. Check that the value lies in the domain.

Answer: (d) \( \frac12 \)

Explanation:
1. \( \cos^{-1}(\sqrt3x)=\frac\pi2-\cos^{-1}x=\sin^{-1}x \).
2. Taking cosines: \( \sqrt3x=\cos(\sin^{-1}x)=\sqrt{1-x^2} \).
3. \( 3x^2=1-x^2 \), so \( x=\frac12 \) (x must be positive).

Teacher's Note:
1. Squaring may introduce extra roots; reject x = -1/2 because cos⁻¹ of a negative number would then exceed π/2 and violate the equation.

Answer: (a) 13

Explanation:
1. The identity \( \sin^{-1}\frac ax+\sin^{-1}\frac bx=\frac\pi2 \) gives \( x^2=a^2+b^2 \).
2. \( x^2=25+144=169 \), so x = 13.

Teacher's Note:
1. The numbers 5, 12, 13 form a Pythagorean triple. The two angles are the acute angles of the right triangle.

Answer: (c) 3

Explanation:
1. \( \operatorname{cosec}^{-1}\frac54=\sin^{-1}\frac45 \).
2. \( \sin^{-1}\frac x5=\frac\pi2-\sin^{-1}\frac45=\cos^{-1}\frac45=\sin^{-1}\frac35 \).
3. So x = 3.

Teacher's Note:
1. Convert cosec⁻¹ to sin⁻¹ and use the complementary angle.

Answer: (d) \( \frac\pi4+\frac x2 \)

Explanation:
1. \( 1+\sin x=\left(\cos\frac x2+\sin\frac x2\right)^2 \) and \( \cos x=\cos^2\frac x2-\sin^2\frac x2 \).
2. The ratio is \( \frac{\cos\frac x2+\sin\frac x2}{\cos\frac x2-\sin\frac x2}=\tan\left(\frac\pi4+\frac x2\right) \).
3. The angle \( \frac\pi4+\frac x2 \) lies in \( (0,\frac\pi2) \), so the value is \( \frac\pi4+\frac x2 \).

Teacher's Note:
1. Use half-angle forms of 1 + sin x and cos x. The range of x ensures the result is in the principal range.

Answer: (c) \( \frac\pi6 \)

Explanation:
1. If \( \cos\theta=\sqrt{\frac23} \), then \( \sin\theta=\sqrt{\frac13} \).
2. \( \tan^{-1}\frac1{\sqrt3}=\frac\pi6 \).

Teacher's Note:
1. A right triangle with adjacent √2 and hypotenuse √3 gives the opposite side 1.

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

Explanation:
1. \( \sin^{-1}\frac{2a}{1+a^2}=2\tan^{-1}a \) and \( \cos^{-1}\frac{1-a^2}{1+a^2}=2\tan^{-1}a \) for \( 0\le a<1 \).
2. The bracket is \( \tan^{-1}a+\tan^{-1}a=2\tan^{-1}a \).
3. \( \tan(2\tan^{-1}a)=\frac{2a}{1-a^2} \).

Teacher's Note:
1. The standard conversions need 0 ≤ a ≤ 1. The result is then the double-angle formula.

Answer: (d) \( \frac12 \)

Explanation:
1. \( \tan^{-1}(2-\sqrt3)=\frac\pi{12} \), so \( 3\tan^{-1}(2-\sqrt3)=\frac\pi4 \).
2. \( \tan^{-1}x=\frac\pi4-\tan^{-1}\frac13=\tan^{-1}\frac{1-\frac13}{1+\frac13}=\tan^{-1}\frac12 \).
3. So \( x=\frac12 \).

Teacher's Note:
1. Recognise 2 - √3 = tan 15 degrees. Then isolate tan⁻¹ x.

Answer: (b) \( \frac{220}{221} \)

Explanation:
1. Let \( A=\cos^{-1}\frac5{13} \), \( B=\cos^{-1}\frac8{17} \). Since 5/13 < 8/17, A > B so A - B is positive.
2. \( \cos(A-B)=\frac5{13}\cdot\frac8{17}+\frac{12}{13}\cdot\frac{15}{17}=\frac{40+180}{221}=\frac{220}{221} \).
3. A - B is in [0, π], so \( x=\frac{220}{221} \).

Teacher's Note:
1. The order of the terms matters: the difference must be non-negative to equal a principal cos⁻¹ value. Compare the cosines to decide the order.

Answer: (a) \( 0\le x\le1 \)

Explanation:
1. Let \( \theta=\cos^{-1}x \). Then \( \cos2\theta=2x^2-1 \).
2. \( \cos^{-1}(\cos2\theta)=2\theta \) only when \( 0\le2\theta\le\pi \).
3. So \( 0\le\cos^{-1}x\le\frac\pi2 \), i.e. \( 0\le x\le1 \).

Teacher's Note:
1. The identity cos⁻¹(cos t) = t needs t in [0, π]. For x < 0 the equation fails.

Answer: (c) n

Explanation:
1. Each \( \cos^{-1}x_r\ge0 \), so the sum is zero only if every term is zero.
2. Hence \( x_r=1 \) for all r, and the sum is n.

Teacher's Note:
1. A sum of non-negative quantities equal to zero forces each to be zero.

Answer: (d) \( \sin^{-1}x-\sin^{-1}\sqrt x \)

Explanation:
1. Use \( \sin^{-1}p-\sin^{-1}q=\sin^{-1}\left[p\sqrt{1-q^2}-q\sqrt{1-p^2}\right] \) with p = x, \( q=\sqrt x \).
2. \( p\sqrt{1-q^2}=x\sqrt{1-x} \) and \( q\sqrt{1-p^2}=\sqrt x\sqrt{1-x^2} \).
3. So the value is \( \sin^{-1}x-\sin^{-1}\sqrt x \) (valid because the difference is in the principal range).

Teacher's Note:
1. Match the expression with the sine of a difference. For 0 ≤ x ≤ 1 we have x ≤ sqrt x so the difference is non-positive and in range.

Answer: (a) \( \pi-\theta \)

Explanation:
1. \( \sin\theta=\sin(\pi-\theta) \).
2. For \( \theta\in[\frac\pi2,\frac{3\pi}2] \), \( \pi-\theta\in[-\frac\pi2,\frac\pi2] \), the principal range.
3. So the value is \( \pi-\theta \).

Teacher's Note:
1. The reflected angle π - θ lies in the principal range exactly for this interval.

Answer: (b) \( 2\tan^{-1}\sqrt{\frac ba} \)

Explanation:
1. Let \( t^2=\frac ba \). Then \( \frac{1-t^2}{1+t^2}=\frac{a-b}{a+b} \).
2. \( \cos^{-1}\frac{1-t^2}{1+t^2}=2\tan^{-1}t \) for \( t\ge0 \).
3. So the value is \( 2\tan^{-1}\sqrt{b/a} \).

Teacher's Note:
1. Divide the numerator and the denominator by a to see the pattern (1 - t²)/(1 + t²).

Answer: (c) \( \frac\pi2 \)

Explanation:
1. Since ab = -1, \( \tan(\theta-\phi)=\frac{a-b}{1+ab} \) is undefined (denominator zero).
2. So \( \theta-\phi=\pm\frac\pi2 \).
3. With a > 0 and b < 0, θ is in (0, π/2) and φ in (-π/2, 0), so the difference is positive: \( \frac\pi2 \).

Teacher's Note:
1. When 1 + ab = 0 the tangent of the difference is infinite. The signs of a and b fix which right angle it is.

Answer: (a) \( \left[\frac\pi4,\frac{3\pi}4\right] \)

Explanation:
1. The function is defined for \( x\in[-1,1] \).
2. \( \sin^{-1}x+\cos^{-1}x=\frac\pi2 \), so the value is \( \frac\pi2+\tan^{-1}x \).
3. tan⁻¹ is increasing: it ranges from -π/4 to π/4, so the range is \( [\frac\pi4,\frac{3\pi}4] \).

Teacher's Note:
1. First fix the domain from the first two terms. Then the monotonic term gives the range with both end points included.