JEE Mathematics Inverse Trigonometric Functions MCQs Set 05

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

Explanation:
1. Take \( \tan^{-1}x+\tan^{-1}y=\frac\pi2-\tan^{-1}z=\tan^{-1}\frac1z \).
2. \( \frac{x+y}{1-xy}=\frac1z \), so \( z(x+y)=1-xy \).
3. \( xy+yz+zx=1 \).

Teacher's Note:
1. The reciprocal appears because π/2 minus an angle gives its cotangent. Cross-multiply to get the symmetric result.

Answer: (a) \( \left[-\frac1{\sqrt2},\frac1{\sqrt2}\right] \)

Explanation:
1. The right side lies in \( [-\frac\pi2,\frac\pi2] \), so we need \( |2\sin^{-1}x|\le\frac\pi2 \).
2. That means \( |\sin^{-1}x|\le\frac\pi4 \), i.e. \( |x|\le\frac1{\sqrt2} \).

Teacher's Note:
1. The left side must itself lie in the principal range of sin⁻¹, which limits the angle to π/4.

Answer: (a) \( x=n\pi,\ n\in\mathbb Z \)

Explanation:
1. \( \operatorname{cosec}^{-1}t \) is defined only for \( |t|\ge1 \).
2. \( |\cos x|\le1 \), so we need \( \cos x=\pm1 \).
3. This happens at \( x=n\pi \).

Teacher's Note:
1. Combine the domain of cosec⁻¹ with the range of cosine. Only the boundary value qualifies.

Answer: (d) \( All of the above \)

Explanation:
1. Put \( x=\cos t \) with \( t\in(0,\frac\pi2) \).
2. \( \frac{\sin t}{1+\cos t}=\tan\frac t2 \), so the expression is \( \frac t2=\frac12\cos^{-1}x \).
3. Also \( \cos\frac t2=\sqrt{\frac{1+x}2} \) and \( \sin\frac t2=\sqrt{\frac{1-x}2} \), so all three are equal.

Teacher's Note:
1. A trigonometric substitution x = cos t makes the half-angle identity visible. Check at x = 1/2: all three equal π/6.

Answer: (c) \( [-1,1] \)

Explanation:
1. Both functions are defined exactly for x in [-1, 1].
2. So the sum is defined on [-1, 1].

Teacher's Note:
1. The domain of a sum is the intersection of the individual domains. Both are [-1, 1].

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

Explanation:
1. The logarithm needs \( \sin^{-1}x>0 \).
2. \( \sin^{-1}x>0 \) holds for \( 0<x\le1 \).

Teacher's Note:
1. Both conditions must hold: x in [-1, 1] for sin⁻¹ and sin⁻¹ x > 0 for the logarithm.

Answer: (b) \( \left(-\frac\pi2,\frac\pi2\right) \)

Explanation:
1. The principal value branch of \( \tan^{-1}x \) takes values in \( (-\frac\pi2,\frac\pi2) \), with the end points excluded.

Teacher's Note:
1. tan x is a bijection from (-π/2, π/2) to R. Its inverse has the open interval as range.

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

Explanation:
1. \( \cos\frac{7\pi}6=-\frac{\sqrt3}2 \).
2. \( \cos^{-1}\left(-\frac{\sqrt3}2\right)=\pi-\frac\pi6=\frac{5\pi}6 \).

Teacher's Note:
1. 7 π/6 lies outside [0, π], so use the symmetric angle 2 π - 7 π/6 = 5 π/6.

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

Explanation:
1. \( \sin^{-1}\frac{\sqrt3}2=\frac\pi3 \) and \( \tan^{-1}(-\sqrt3)=-\frac\pi3 \).
2. Value = \( \frac\pi3+\frac\pi3=\frac{2\pi}3 \).

Teacher's Note:
1. tan⁻¹ is an odd function, so tan⁻¹(-√3) = -π/3. Subtracting a negative adds.

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

Explanation:
1. \( \cot\frac\pi{12}=2+\sqrt3 \) since \( \tan15^\circ=2-\sqrt3 \).
2. \( \frac\pi{12}\in(0,\pi) \), so \( \cot^{-1}(2+\sqrt3)=\frac\pi{12} \).

Teacher's Note:
1. Recall tan 15 degrees = 2 - √3, whose reciprocal is 2 + √3.

Answer: (d) \( 0 \)

Explanation:
1. If \( \sin\theta=\sqrt{\frac{13}{17}} \), then \( \cos\theta=\frac2{\sqrt{17}} \) and \( \cot\theta=\frac2{\sqrt{13}} \).
2. \( \sin(\tan^{-1}\frac23)=\frac2{\sqrt{13}} \).
3. The difference is 0.

Teacher's Note:
1. Use right triangles for both terms. The two values happen to be equal.

Answer: (d) 15

Explanation:
1. \( \sec^2\theta=1+\tan^2\theta \): with \( \cot\theta=\frac12 \), \( \tan\theta=2 \), so it equals 5.
2. \( \operatorname{cosec}^2\phi=1+\cot^2\phi \): with \( \tan\phi=\frac13 \), it equals 10.
3. Total = 15.

Teacher's Note:
1. Use sec² = 1 + tan² and cosec² = 1 + cot². Note how the inverse functions flip the argument.

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

Explanation:
1. \( \sin(A+B)=\sin A\cos B+\cos A\sin B \) with \( \sin A=\frac45,\cos A=\frac35,\sin B=\frac7{25},\cos B=\frac{24}{25} \).
2. \( =\frac45\cdot\frac{24}{25}+\frac35\cdot\frac7{25}=\frac{96+21}{125}=\frac{117}{125} \).

Teacher's Note:
1. Find both cosines from the Pythagorean triples. All angles lie in the first quadrant.

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

Explanation:
1. \( \cos2\theta=2\cos^2\theta-1 \) with \( \cos\theta=\frac7{25} \).
2. \( =2\cdot\frac{49}{625}-1=\frac{98-625}{625}=-\frac{527}{625} \).

Teacher's Note:
1. The angle 2 θ is obtuse since cos θ is small, so a negative result is expected.

Answer: (a) \( \frac\pi2 \)

Explanation:
1. \( x+\frac1x=2 \) gives \( (x-1)^2=0 \), so x = 1.
2. \( \sin^{-1}1=\frac\pi2 \).

Teacher's Note:
1. The equation x + 1/x = 2 has a repeated root. Only x = 1 is possible.

Answer: (a) \( -\frac7{17} \)

Explanation:
1. \( \tan\left(2\tan^{-1}\frac15\right)=\frac{2/5}{1-1/25}=\frac5{12} \).
2. \( \tan\left(A-\frac\pi4\right)=\frac{\frac5{12}-1}{1+\frac5{12}}=\frac{-7/12}{17/12}=-\frac7{17} \).

Teacher's Note:
1. First get tan of the double angle. Then apply the subtraction formula with tan (π/4) = 1.

Answer: (a) \( \frac\pi2 \)

Explanation:
1. \( 2\tan^{-1}\frac12=\tan^{-1}\frac{1}{1-\frac14}=\tan^{-1}\frac43 \).
2. \( \sin^{-1}\frac35=\tan^{-1}\frac34 \).
3. \( \tan^{-1}\frac43+\tan^{-1}\frac34=\frac\pi2 \) since the product of the arguments is 1.

Teacher's Note:
1. Convert everything to tan⁻¹. Reciprocal arguments of positive numbers always add to π/2.

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

Explanation:
1. The product 2 x 3 = 6 is greater than 1, so \( \tan^{-1}2+\tan^{-1}3=\pi+\tan^{-1}\frac{5}{1-6}=\pi-\tan^{-1}1 \).
2. \( =\pi-\frac\pi4=\frac{3\pi}4 \).

Teacher's Note:
1. When xy > 1 and x, y > 0, add π to the principal value. Both angles exceed 45 degrees, so the sum is beyond 90 degrees.

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

Explanation:
1. \( \tan(A-B)=\frac{\frac mn-\frac{m-n}{m+n}}{1+\frac mn\cdot\frac{m-n}{m+n}} \).
2. Numerator = \( \frac{m(m+n)-n(m-n)}{n(m+n)}=\frac{m^2+n^2}{n(m+n)} \); denominator = \( \frac{n(m+n)+m(m-n)}{n(m+n)}=\frac{m^2+n^2}{n(m+n)} \).
3. The ratio is 1, so the difference is \( \frac\pi4 \).

Teacher's Note:
1. Both numerator and denominator simplify to the same expression. Check with m = 2, n = 1.

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

Explanation:
1. \( 2\tan^{-1}\frac13=\tan^{-1}\frac{2/3}{1-1/9}=\tan^{-1}\frac34 \).
2. \( \tan^{-1}\frac34+\tan^{-1}\frac17=\tan^{-1}\frac{\frac34+\frac17}{1-\frac3{28}}=\tan^{-1}\frac{25/28}{25/28}=\frac\pi4 \).

Teacher's Note:
1. Combine step by step. The product of the arguments is below 1, so no π correction is needed.

Answer: (c) 5/3

Explanation:
1. \( \tan^{-1}5+\tan^{-1}\frac15=\frac\pi2 \).
2. The bracket is \( \frac\pi2-\tan^{-1}\frac34=\tan^{-1}\frac43 \).
3. \( \sec\left(\tan^{-1}\frac43\right)=\sqrt{1+\frac{16}9}=\frac53 \).

Teacher's Note:
1. Reciprocal pair arguments give π/2. Then use a 3-4-5 triangle.

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

Explanation:
1. \( 2\tan^{-1}\frac15=\tan^{-1}\frac5{12} \) and \( 4\tan^{-1}\frac15=\tan^{-1}\frac{120}{119} \).
2. \( \tan^{-1}\frac{120}{119}-\frac\pi4=\tan^{-1}\frac1{239} \), so \( 4\tan^{-1}\frac15=\frac\pi4+\tan^{-1}\frac1{239} \).
3. \( \tan^{-1}\frac1{99}-\tan^{-1}\frac1{70}=\tan^{-1}\frac{-29/6930}{6931/6930}=-\tan^{-1}\frac{29}{6931}=-\tan^{-1}\frac1{239} \).
4. The two extra terms cancel, leaving \( \frac\pi4 \).

Teacher's Note:
1. This is a Machin-type identity. Numerical evaluation is a reliable check: 0.7854 equals π/4.

Answer: (b) \( [-2,1] \)

Explanation:
1. We need \( -1\le\frac{2x+1}3\le1 \).
2. \( -3\le2x+1\le3 \), so \( -2\le x\le1 \).

Teacher's Note:
1. Multiply the inequality by 3 and subtract 1 before dividing by 2. Both ends are included.

Answer: (c) \( \left[0,\frac12\right] \)

Explanation:
1. The square root needs \( x\ge0 \).
2. \( 0\le\sqrt{2x}\le1 \) gives \( 2x\le1 \).
3. So \( 0\le x\le\frac12 \).

Teacher's Note:
1. Both the square root and the cos⁻¹ impose conditions. Combine them.

Answer: (d) \( (0,\pi) \)

Explanation:
1. The principal branch of cot⁻¹ takes values strictly between 0 and π.

Teacher's Note:
1. cot x is a bijection from (0, π) to R, so its inverse has the open interval (0, π) as range.

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

Explanation:
1. \( \sin\frac{2\pi}3=\sin\left(\pi-\frac{2\pi}3\right)=\sin\frac\pi3 \).
2. \( \frac\pi3 \) lies in \( [-\frac\pi2,\frac\pi2] \), so the value is \( \frac\pi3 \).

Teacher's Note:
1. Since 2 π/3 is outside the principal range, reflect it about π/2.

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

Explanation:
1. \( \tan\left(-\frac{5\pi}4\right)=-\tan\frac{5\pi}4=-\tan\frac\pi4=-1 \).
2. \( \sin^{-1}(-1)=-\frac\pi2 \).

Teacher's Note:
1. Tangent has period π, so 5 π/4 behaves like π/4. The negative sign carries through.