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[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).