Class 12 - Maths - Inverse Trigonometric Functions

Exercise 2.1

Question 1:

Find the principal value of sin−1(−1/2)

Answer:

Let sin−1(−1/2) = y, then

sin y = -1/2 = -sin(π/6) = sin(-π/6)

We know that the range of the principal value branch of sin-1 is [-π/2, π/2]

and sin(−π/6) = -1/2

Hence, the principal value of sin−1(−1/2) = −π/6.

Question 2:

Find the principal value of cos−1(√3/2)

Answer:

Let cos−1(√3/2) = y then

cos y = √3/2 = cos(π/6)

We know that the range of the principal value branch of cos-1 is [0, π]

and cos(π/6) = √3/2

Hence, the principal value of cos−1(√3/2) = π/6.

Question 3:

Find the principal value of cosec−1(2)

Answer:

Let cosec−1(2) = y, then

cosec y = 2 = cosec(π/6)

We know that the range of the principal value branch of cosec-1 is [-π/2, π/2] – {0}

and cosec(π/6) = 2

Hence, the principal value of cosec−1(2) = π/6.

Question 4:

Find the principal value of tan−1(-√3)

Answer:

Let tan−1(-√3) = y, then

tan y = -√3 = -tan(π/3) = tan(-π/3)

We know that the range of the principal value branch of tan-1 is [-π/2, π/2]

and tan(-π/3) = -√3

Hence, the principal value of tan−1(-√3) = -π/3.

Question 5:

Find the principal value of cos−1(-1/2)

Answer:

Let cos−1(-1/2) = y then

cos y = -1/2 = -cos(π/3) = cos(π - π/3) = cos(2π/3)

We know that the range of the principal value branch of cos-1 is [0, π]

and cos(2π/3) = -1/2

Hence, the principal value of cos−1(-1/2) = 2π/3.

Question 6:

Find the principal value of tan−1(-1)

Answer:

Let tan−1(-1) = y, then

tan y = -1 = -tan(π/4) = tan(-π/4)

We know that the range of the principal value branch of tan-1 is [-π/2, π/2]

and tan(-π/4) = -1

Hence, the principal value of tan−1(-1) = -π/4.

Question 7:

Find the principal value of sec−1(2/√3)

Answer:

Let sec−1(2/√3) = y, then

sec y = 2/√3 = sec(π/6)

We know that the range of the principal value branch of tan-1 is [0, π] – { π/2}

and sec(π/6) = 2/√3

Hence, the principal value of sec−1(2/√3) = π/6.

Question 8:

Find the principal value of cot−1(√3)

Answer:

Let cot−1(√3) = y, then

cot y = √3 = cot(π/6)

We know that the range of the principal value branch of tan-1 is (0, π]

and cot(π/6) = √3

Hence, the principal value of cot−1(√3) = π/6.

Question 9:

Find the principal value of cos−1(-1/√2)

Answer:

Let cos−1(-1/√2) = y then

cos y = -1/√2 = -cos(π/4) = cos(π - π/4) = cos(3π/4)

We know that the range of the principal value branch of cos-1 is [0, π]

and cos(3π/4) = -1/√2

Hence, the principal value of cos−1(-1/√2) = 3π/4.

Question 10:

Find the principal value of cosec−1(-√2)

Answer:

Let cosec−1(-√2) = y, then

cosec y = -√2 = -cosec(π/4) = cosec(-π/4)

We know that the range of the principal value branch of cosec-1 is [-π/2, π/2] – {0}

and cosec(-π/4) = -√2

Hence, the principal value of cosec−1(-√2) = -π/4.

Question 11:

Find the value of tan−1(1) + cos−1(-1/2) + sin−1(-1/2)

Answer:

Let tan−1(1) = y, then

tan y = 1 = tan π/4

We know that the range of the principal value branch of tan-1 is (-π/2, π/2)

So, tan−1(1) = π/4

Let cos−1(-1/2) = y, then

cos y = -1/2 = -cos π/3 = cos(π – π/3) =cos 2π/3

We know that the range of the principal value branch of cos-1 is [0, π]

So, cos−1(-1/2) = 2π/3

Let sin−1(-1/2) = y, then

sin y = -1/2 = -sin π/6 = sin(-π/6)

We know that the range of the principal value branch of sin-1 is [-π/2, π/2]

So, sin−1(-1/2) = -π/6

Now, tan−1(1) + cos−1(-1/2) + sin−1(-1/2) = π/4 + 2π/3 - π/6

= (3π + 8π - 2π)/12

= 9π/12

= 3π/4

Question 12:

Find the value of cos−1(1/2) + 2sin−1(1/2)

Answer:

Let cos−1(1/2) = y, then

cos y = 1/2 = cos π/3

We know that the range of the principal value branch of cos-1 is [0, π]

So, cos−1(1/2) = π/3

Let sin−1(1/2) = y, then

sin y = 1/2 = sin π/6

We know that the range of the principal value branch of sin-1 is [-π/2, π/2]

So, sin−1(1/2) = π/6

Now, cos−1(1/2) + 2sin−1(1/2) = π/3 + 2π/6

= π/3 + π/3

= 2π/3

Question 13:

If sin-1 x = y, then

(A) 0 ≤ y < π            (B) – π/2 ≤ y ≤ π/2             (C) 0 < y < π            (D) – π/2 < y < π/2

Answer:

It is given, sin-1 x = y

We know that the range of the principal value branch of sin-1 is [-π/2, π/2]

So, – π/2 ≤ y ≤ π/2

Hence, option (B) is the correct answer.

Question 14:

tan-1 √3 – sec-1(-2) is equal to

(A) π                                    (B) – π/3                              (C) π/3                          (D) 2π/3

Answer:

Let tan−1(√3) = y, then

tan y = √3 = tan π/3

We know that the range of the principal value branch of tan-1 is (-π/2, π/2)

So, tan−1(√3) = π/3

Let sec−1(-2) = y, then

sec y = -2 = -sec π/3 = sec(π – π/3) =sec 2π/3

We know that the range of the principal value branch of sec-1 is [0, π] – {π/2}

So, sec−1(-2) = 2π/3

Now, tan−1(√3) - sec−1(-2) = π/3 - 2π/3

= -π/3

Hence, option (B) is the correct answer.

Exercise 2.2

Question 1:

Prove that 3sin-1 x = sin-1(3x – 4x3), x Є [-1/2, 1/2]

Answer:

Let sin-1 x = y, then sin y = x

RHS:

sin-1(3x – 4x3) = sin-1(3sin y – 4sin3 y)

= sin-1(sin 3y)

= 3y

= 3 sin-1 x

= LHS

Question 2:

Prove that 3 cos-1 x = cos-1(4x3 – 3x), x Є [1/2, 1]

Answer:

Let cos-1 x = y, then cos y = x

RHS:

cos-1(4x3 – 3x) = cos-1(4cos3 y – 3cos y)

= cos-1(cos 3y)

= 3y

= 3 cos-1 x

= LHS

Question 3:

Prove that: tan-1 (2/11) + tan-1 (7/24) = tan-1 (1/2)

Answer:

LHS:

Given, tan-1 (2/11) + tan-1 (7/24)

= tan-1 {(2/11 + 7/24)/(1 - 2/11 * 7/24)}

= tan-1 [{(48 + 77)/(11 * 24)}/{(11 * 24 - 14)/(11 * 24)}]

= tan-1 {(48 + 77)/(264 - 14)}

= tan-1 (125/250)

= tan-1 (1/2)

= RHS

So, tan-1 (2/11) + tan-1 (7/24) = tan-1 (1/2)

Question 4:

Prove that: 2tan-1 (1/2) + tan-1 (1/7) = tan-1 (31/17)

Answer:

LHS:

Given, 2tan-1 (1/2) + tan-1 (1/7)

= tan-1 [(2 * 1/2)/{1 – (1/2)2} + tan-1 (1/7)

= tan-1 [1/(1 – 1/4)] + tan-1 (1/7)

= tan-1 [1/(3/4)] + tan-1 (1/7)

= tan-1 (4/3) + tan-1 (1/7)

= tan-1 {(4/3 + 1/7)/(1 – 4/3 * 1/7)}

= tan-1 {(4/3 + 1/7)/(1 – 4/21)}

= tan-1 [{(28 + 3)/21}/{(21 - 4)/21}]

= tan-1 (31/17)

= RHS

So, 2tan-1 (1/2) + tan-1 (1/7) = tan-1 (31/17)

Question 5:

Write the function tan-1{√(1 + x2) - 1}/x, x ≠ 0, in the simplest form.

Answer:

Let x = tan y

Now, tan-1{√(1 + x2) - 1}/x = tan-1[{√(1 + tan2 y) - 1}/tan y]

= tan-1[(sec y – 1)/tan y]

= tan-1[(1 – cos y)/sin y]

= tan-1[(2sin2 y/2)/(2 * sin y/2  * cos y/2)]

= tan-1[tan y/2]

= y/2

= (tan-1 x)/2

Question 6:

Write the function tan-1{1/√(x2 – 1)}, |x| > 1, in the simplest form.

Answer:

Let x = cosec y

Now, tan-1{1/√(1 + x2)} = tan-1{1/√(cosec2 y – 1)}

= tan-1[1/cot y]

= tan-1[tan y]

= y

= cosec-1 x

= π/2 - sec-1 x

Question 7:

Write the function tan-1[√{(1 – cos x)/(1 + cos x)}], x < π in the simplest form.

Answer:

Given function is tan-1[√{(1 – cos x)/(1 + cos x)}]

Now, tan-1[√{(1 – cos x)/(1 + cos x)}] = tan-1[√{(2sin2 x/2)/ (2cos2 x/2)}]

= tan-1[√(tan2 x/2)]

= tan-1[tan x/2]

= x/2

Question 8:

Write the function tan-1[(cos x –sin x)/(cos x + sin x)], 0 < x < π in the simplest form.

Answer:

Given function is tan-1[(cos x –sin x)/(cos x + sin x)]

Now, tan-1[(cos x –sin x)/(cos x + sin x)] = tan-1[(1 – sin x/cos x)/(1 + sin x/cos x)]

= tan-1[(1 – tan x)/(1 + tan x)]

= tan-1[(tan π/4 – tan x)/(1 + tan π/4 * tan x)]

= tan-1[tan (π/4 – x)]

= π/4 – x

Question 9:

Write the function tan-1[x/√(a2 – x2)], |x| < a in the simplest form.

Answer:

Let x = a sin y

Now, tan-1[x/√(a2 – x2)] = tan-1[a sin y/√(a2 – a2 sin2 y)]

= tan-1[a sin y/{a√ (1 – sin2 y)}]

= tan-1[(a sin y)/(a cos y)]

= tan-1(tan y)

= y

= sin-1 (x/a)

Question 10:

Write the function in tan-1[(3a2x – x3)/(a3 – 3ax2)], a > 0, -a/√3 ≤ x ≤ a/√3 the simplest form.

Answer:

Let x = a tan y

Now, tan-1[(3a2x – x3)/(a3 – 3ax2)] = tan-1[(3a2 * a tan y – a3 tan3 y)/(a3 – 3a * a2 tan2 y)]

= tan-1[(3a3 tan y – a3 tan3 y)/(a3 – 3a3 tan2 y)]

= tan-1[(3 tan y – tan3 y)/(1 – 3 tan2 y)]

= tan-1[tan 3y]

= 3y

= 3tan-1(x/a)

Question 11:

Find the value of tan-1[2 cos(2 sin-1 1/2)]

Answer:

Given, tan-1[2 cos(2 sin-1 1/2)] = tan-1[2 cos(2 sin-1 (sin π/6))]

= tan-1[2 cos(2 * π/6)]

= tan-1[2 cos(π/3)]

= tan-1[2 * 1/2]

= tan-1

= π/4

Question 12:

Find the value of cot(tan-1 a + cot-1 a)

Answer:

Given, cot(tan-1 a + cot-1 a) = cot(π/2) = 0                     [Since tan-1 x + cot-1 x = π/2]

Question 13:

Find the value of tan [sin-1{2x/(1 + x2)} + cos-1{(1 – y2)/(1 + y2)}]/2, |x| < 1, y > 0 and xy < 1

Answer:

Given, tan [sin-1{2x/(1 + x2)} + cos-1{(1 – y2)/(1 + y2)}]/2

= tan [2 tan-1 x + 2 tan-1 y]/2

= tan [tan-1 x + tan-1 y]

= tan [tan-1 {(x + y)/(1 - xy)}]

= (x + y)/(1 - xy)

Question 14:

If sin(sin-1 1/5 + cos-1 x) = 1, then find the value of x.

Answer:

Given, sin(sin-1 1/5 + cos-1 x) = 1

=> sin-1 1/5 + cos-1 x = sin-1 1

=> sin-1 1/5 + cos-1 x = π/2

=> sin-1 1/5 = π/2 - cos-1 x                     [Since sin-1 x + cos-1 x = π/2]

=> sin-1 1/5 = sin-1 x

=> x = 1/5

Question 15:

If tan-1 {(x - 1)/(x - 2)} + tan-1 {(x + 1)/(x + 2)} = π/4, then find the value of x.

Answer:

Given, tan-1 {(x - 1)/(x - 2)} + tan-1 {(x + 1)/(x + 2)} = π/4

=> tan-1 [{(x - 1)/(x - 2) + (x + 1)/(x + 2)}/{1 - (x - 1)/(x - 2) * (x + 1)/(x + 2)}] = π/4

=> {(x - 1)/(x - 2) + (x + 1)/(x + 2)}/{1 - (x - 1)/(x - 2) * (x + 1)/(x + 2)} = tan π/4

=> {(x - 1)/(x - 2) + (x + 1)/(x + 2)}/{1 - (x - 1)/(x - 2) * (x + 1)/(x + 2)} = 1

=> [{(x - 1) * (x + 2) + (x + 1)/(x - 2)}/{(x - 2)(x + 2)}]    = 1

[{(x - 2)(x + 2) - (x - 1)(x + 1)}/{(x - 2)/(x + 2)}]

=> {(x - 1) * (x + 2) + (x + 1)(x - 2)}/{(x - 2)(x + 2) - (x - 1)(x + 1)} = 1

=> (x2 + 2x – x – 2 + x2 + x – 2x - 2)/{x2 – 4 – (x2 - 1)} = 1

=> (2x2 – 4)/(-3) = 1

=> 2x2 – 4 = -3

=> 2x2 = -3 + 4

=> 2x2 = 1

=> x2 = 1/2

=> x = ±1/√2

So, the value of x is ±1/√2

Question 16:

Find the value of sin-1(sin 2π/3)

Answer:

Given, sin-1(sin 2π/3)

We know that sin-1(sin x) = x if x Є [-π/2, π/2], which is the principal value branch of sin-1 x.

So, sin-1(sin 2π/3) = sin-1(sin {π - π/3}) = sin-1(sin π/3) = π/3 Є [-π/2, π/2]

Hence, sin-1(sin 2π/3) = π/3

Question 17:

Find the value of tan-1(tan 3π/4)

Answer:

Given, tan-1(tan 3π/4)

We know that tan-1(tan x) = x if x Є [-π/2, π/2], which is the principal value branch of tan-1 x.

So, tan-1(tan 3π/4) = tan-1(tan {π - π/4}) = tan-1(-tan π/4) = tan-1(tan {-π/4}) = -π/4 Є [-π/2, π/2]

Hence, tan-1(tan 3π/4) = -π/4

Question 18:

Find the value of tan(sin-1 3/5 + cot-1 3/2)

Answer:

Given, tan(sin-1 3/5 + cot-1 3/2) = tan[tan-1 {3/√(52 - 32)} + tan-1 2/3]

= tan(tan-1 3/4 + tan-1 2/3)

= tan[tan-1 {(3/4 + 2/3)/(1 – 3/4 * 2/3)}]

= tan[tan-1 {(17/12)/(1 – 6/12)}]

= tan[tan-1 {(17/12)/(6/12)}]

= tan(tan-1 17/6)

= 17/6

Question 19:

cos-1(cos 7π/6) is equal to

(A) 7π/6                             (B) 5π/6                         (C) π/3                               (D) π/6

Answer:

Given, cos-1(cos 7π/6)

We know that cos-1(cos x) = x if x Є [0, π], which is the principal value branch of cos-1 x.

So, cos-1(cos 7π/6) = cos-1(cos {2π - 5π/6}) = cos-1(cos 5π/6) = 5π/6 Є [0, π]

Hence, cos-1(cos 7π/6) = 5π/6

Hence, the option (B) is the correct answer.

Question 20:

sin(π/3- sin-1 (-1/2))

(A) 1/2                             (B) 1/3                         (C) 1/4                               (D) 1

Answer:

Given, sin(π/3- sin-1 (-1/2))

We know that the range of the principal value branch of sin-1 is [-π/2, π/2]

So, sin(π/3 - sin-1 (-1/2)) = sin[π/3 - sin-1(-sin π/6)]

= sin[π/3 - sin-1(sin (-π/6))]

= sin(π/3 + π/6)

= sin 3π/6

= sin π/2

= 1

Hence, sin(π/3 - sin-1 (-1/2)) = 1

Therefore, the correct option is (D).

Question 21:

tan-1 √3 - cot-1 (-√3) is equal to

(A) π                           (B) - π/2                        (C) 0                                    (D) 2√3

Answer:

Given, tan-1 √3 - cot-1 (-√3)

We know that the range of the principal value branch of tan-1 is (-π/2, π/2) and cot-1 is (0, π).

So, tan-1 √3 - cot-1 (-√3) = tan-1 (tan π/3) - cot-1 (-cot π/6)

= π/3 - cot-1 [cot (π - π/6)]

= π/3 - cot-1 (cot 5π/6)

= π/3 - 5π/6

= (2π - 5π)/6

= -3π/6

= -π/2

Hence, tan-1 √3 - cot-1 (-√3) = -π/2

Therefore, the correct option is (B).

Miscellaneous Exercise

Question 1:

Find the value of cos-1(cos 13π/6).

Answer:

Given, cos-1(cos 13π/6)

We know that cos-1(cos x) = x if x Є [0, π], which is the principal value branch of cos-1 x.

So, cos-1(cos 13π/6) = cos-1(cos {2π + π/6}) = cos-1(cos π/6) = π/6 Є [0, π]

Hence, cos-1(cos 13π/6) = π/6

Question 2:

Find the value of tan-1(tan 7π/6).

Answer:

Given, tan-1(tan 7π/6)

We know that tan-1(tan x) = x if x Є (-π/2, π/2), which is the principal value branch of tan-1 x.

So, tan-1(tan 7π/6) = tan-1(tan {π + π/6}) = tan-1(tan π/6) = π/6 Є (-π/2, π/2)

Hence, tan-1(tan 7π/6) = π/6

Question 3:

Prove that 2sin-1 3/5 = tan-1 24/7

Answer:

LHS:

Given, 2sin-1 3/5 = 2tan-1 {3/√(52 - 32)}

= 2tan-1 3/4

= tan-1{(2 * 3/4)/(1 – (3/4)2)}

= tan-1{(3/2)/(1 – 9/16)}

= tan-1{(3/2)/(7/16)}

= tan-1{(3/2) * (16/7)}

= tan-1 24/7

= RHS

Question 4:

Prove that sin-1 8/17 + sin-1 3/5 = tan-1 77/36

Answer:

LHS:

Given, sin-1 8/17 + sin-1 3/5 = tan-1 {8/√(172 - 82)} + tan-1 {3/√(52 - 32)}

= tan-1 8/15 + tan-1 3/4

= tan-1 {(8/15 + 3/4)/(1 – 8/15 * 3/4)}

= tan-1 {(77/60)/(1 – 24/60)}

= tan-1 {(77/60)/(36/60)}

= tan-1 (77/36)

= RHS

Question 5:

Prove that cos-1 4/5 + cos-1 12/13 = cos-1 33/65

Answer:

LHS:

Given, cos-1 4/5 + cos-1 12/13 = tan-1 {√(52 - 42)/4} + tan-1 {√(132 - 122)/12}

= tan-1 3/4 + tan-1 5/12

= tan-1 {(3/4 + 5/12)/(1 – 3/4 * 5/12)}

= tan-1 {(14/12)/(48/33)}

= tan-1 (56/33)

= cos-1 {33/√(562 + 332)}

= cos-1 {33/√4225}

= cos-1 33/65

= RHS

Question 6:

Prove that cos-1 12/13 + sin-1 3/5 = sin-1 56/65

Answer:

LHS:

Given, cos-1 12/13 + sin-1 3/5 = tan-1 {√(132 - 122)/12} + tan-1 {3/√(52 - 32)}

= tan-1 5/12 + tan-1 3/4

= tan-1 {(5/12 + 3/4)/(1 – 3/4 * 5/12)}

= tan-1 {(14/12)/(33/48)}

= tan-1 (56/33)

= sin-1 {56/√(562 + 332)}

= sin-1 {56/√4225}

= sin-1 56/65

= RHS

Question 7:

Prove that tan-1 63/16 = sin-1 5/13 + cos-1 3/5

Answer:

RHS:

Given, sin-1 5/13 + cos-1 3/5 = tan-1 {5/√(132 - 52)} + tan-1 {√(52 - 32)/3}

= tan-1 5/12 + tan-1 4/3

= tan-1 {(5/12 + 4/3)/(1 – 4/3 * 5/12)}

= tan-1 {(21/12)/(1 – 20/36)}

= tan-1 {(21/12)/(16/36)}

= tan-1 (63/16)

= LHS

Question 8:

Prove that tan-1 1/5 + tan-1 1/7 + tan-1 1/3 + tan-1 1/8 = π/4

Answer:

LHS:

tan-1 1/5 + tan-1 1/7 + tan-1 1/3 + tan-1 1/8

= tan-1 {(1/5 + 1/7)/(1 – 1/5 * 1/7)} + tan-1 {(1/3 + 1/8)/(1 – 1/3 * 1/8)}

= tan-1 {(12/35)/(1 – 1/35)} + tan-1 {(11/24)/(1 – 1/24)}

= tan-1 {(12/35)/(34/35)} + tan-1 {(11/24)/(23/24)}

= tan-1 12/34 + tan-1 11/23

= tan-1 6/17 + tan-1 11/23

= tan-1 {(6/17 + 11/23)/(1 – 6/17 * 11/23)}

= tan-1 {(325/391)/(1 – 66/391)}

= tan-1 {(325/391)/(325/391)}

= tan-1 1

= π/4

= RHS

Question 9:

Prove that tan-1 √x = (1/2) * cos-1 {(1 - x)/(1 + x)}, x Є [0, 1]

Answer:

LHS:

Given, tan-1 √x = (1/2) * 2 tan-1 √x

= (1/2) * cos-1 {(1 – (√x)2)/(1 + (√x)2)}

= (1/2) * cos-1 {(1 - x)/(1 + x)}                [Since 2tan-1 x = cos-1 {(1 – x2)/(1 + x2)}]

= RHS

Question 10:

Prove that cot-1 [{√(1 + sin x) + √(1 - sin x)}/{√(1 + sin x) - √(1 - sin x)}] = x/2, x Є (0, π/4)

Answer:

LHS:

Given, cot-1 [{√(1 + sin x) + √(1 - sin x)}/{√(1 + sin x) - √(1 - sin x)}]

Now, consider {√(1 + sin x) + √(1 - sin x)}/{√(1 + sin x) - √(1 - sin x)}

= {√(1 + sin x) + √(1 - sin x)}2/[{√(1 + sin x) - √(1 - sin x)} * {√(1 + sin x) + √(1 - sin x)}]

= [(1 + sin x) + (1 - sin x) + 2√{(1 + sin x)(1 – sin x)}]/[{√(1 + sin x)}2 – {√(1 - sin x)}2]

= [2{(1 + √(1 – sin2 x)}]/[(1 + sin x) – (1 - sin x)]

= [2(1 + cos x)]/(2 * sin x)

= (1 + cos x)/sin x

= (2 * cos2 x/2)/(2 * sin x/2 * cos x/2)

= cot x/2

Now, cot-1 [{√(1 + sin x) + √(1 - sin x)}/{√(1 + sin x) - √(1 - sin x)}] = cot-1 [cot x/2]

= x/2

= RHS

Question 11:

Prove that tan-1 [{√(1 + x) - √(1 - x)}/{√(1 + x) + √(1 - x)}] = π/4 - (1/2)cos-1 x, -1/√2 ≤ x ≤ 1

Answer:

LHS:

Given, tan-1 [{√(1 + x) - √(1 - x)}/{√(1 + x) + √(1 - x)}]

= tan-1 [{√(1 + cos y) - √(1 - cos y)}/{√(1 + cos y) + √(1 - cos y)}]               [Let x = cos y]

= tan-1 [{√(2cos2 y/2) - √(2sin2 y/2)}/{√(2cos2 y/2) + √(2sin2 y/2)}]

= tan-1 [{√2cos y/2 - √2sin y/2}/{√2cos y/2 + √2sin y/2}]

= tan-1 [{cos y/2 - sin y/2}/{cos y/2 + sin y/2}]

= tan-1 [{1 - tan y/2}/{1 + tan y/2}]                                       [divide by cos y/2]

= tan-1 [{tan π/4 - tan y/2}/{1 + tan π/4 * tan y/2}]

= tan-1 [tan (π/4 - y/2)]

= π/4 - y/2

= π/4 - (1/2)cos-1 x

= RHS

Question 12:

9π/8 – (9/4)sin-1 1/3 = (9/4)sin-1 2√2/3

Answer:

LHS:

Given, 9π/8 – (9/4)sin-1 1/3

= (9/4)[π/2 – sin-1 1/3]

= (9/4)[cos-1 1/3]                         [since sin-1 x + cos-1 x = π/2]

= (9/4)[sin-1 √(32 – 12)/3]

= (9/4)[sin-1 √8/3]

= (9/4)[sin-1 2√2/3]

= RHS

Question 13:

Solve for x: 2tan-1(cos x) = tan-1(2cosec x)

Answer:

Given, 2tan-1(cos x) = tan-1(2cosec x)

= tan-1{2cos x/(1 – cos2 x)} = tan-1(2cosec x)                 [Since 2tan-1 x = tan-1{2x/(1 – x2)}]

= 2cos x/(1 – cos2 x) = 2cosec x

= 2cos x/sin2 x = 2/sin x

= sin x * cos x = sin2 x

= sin x * cos x - sin2 x = 0

= sin x(cos x - sin x) = 0

= sin x = 0 or cos x − sin x = 0

But sin x ≠ 0 as it does not satisfy the equation

So, cos x − sin x = 0

⇒ cos x = sin x

⇒ tan x = 1

⇒ tan x = tan π/4

⇒ x = π/4

Question 14:

Solve for x: tan-1 {(1 - x)/(1 + x)} = (1/2)tan-1 x, x > 0

Answer:

Given, tan-1 {(1 - x)/(1 + x)} = (1/2)tan-1 x

=> tan-1 1 - tan-1 x = (1/2)tan-1 x                    [Since tan-1 x - tan-1 y = tan-1 {(x - y)/(1 + xy)}]

=> π/4 - tan-1 x = (1/2)tan-1 x

=> π/4 = tan-1 x + (1/2)tan-1 x

=> π/4 = (3/2)tan-1 x

=> tan-1 x = π/6

=> x = tan π/6

=> x = 1/√3

Question 15:

sin(tan-1 x), |x| < 1 is equal to

(A) x/√(1 – x2)                  (B) 1/√(1 – x2)                      (C) 1/√(1 + x2)                  (D) x/√(1 + x2)

Answer:

Given, sin(tan-1 x) = sin(sin-1{x/√(1 + x2)})           [Since tan-1 a/b = sin-1{a/√(a2 + b2)}]

= x/√(1 + x2)

Hence, the correct answer is option (D).

Question 16:

sin-1(1 – x) - 2sin-1 x = π/2, then x is equal to

(A) 0, 1/2                            (B) 1, 1/2                               (C) 0                             (D) 1/2

Answer:

Given, sin-1(1 – x) - 2sin-1 x = π/2

Let x = sin y

So, sin-1(1 – sin y) – 2y = π/2

=> sin-1(1 – sin y) = π/2 + 2y

=> 1 – sin y = sin(π/2 + 2y)

=> 1 – sin y = cos 2y

=> 1 – sin y = 1 – 2sin2 y

=> 2sin2 y – sin y = 0

=> 2x2 – x = 0                                [Since x = sin y]

=> x(2x - 1) = 0

=> x = 0, 1/2

But x = 1/2 does not satisfy the given equation.

So, x = 0 is the solution of the given equation

Hence, the correct answer is option (C).

Question 17:

tan-1  x/y - tan-1{(x – y)/(x + y)} is equal to

(A) π/2                               (B) π/3                         (C) π/4                            (D) -3π/4

Answer:

Given, tan-1  x/y - tan-1{(x – y)/(x + y)}

= tan-1 [{x/y - (x – y)/(x + y)}/{1 – (x/y)  * (x – y)/(x + y)}]

= tan-1 [{x(x + y) - y(x – y)}/{y(x + y)}/ {y(x + y) + x(x – y)}/{y(x + y)}]

= tan-1 [{x(x + y) - y(x – y)}/{y(x + y) + x(x – y)}]

= tan-1 [{x2 + xy - xy + y2}/{xy + y2 + x2 – xy}]

= tan-1 [{x2 +  y2}/{y2 + x2}]

= tan-1 1

= π/4

Hence, the correct answer is option (C).

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