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 – 4x^{3}), x Є [-1/2, 1/2]

Answer:

Let sin^{-1} x = y, then sin y = x

RHS:

sin^{-1}(3x – 4x^{3}) = sin^{-1}(3sin y – 4sin^{3} y)

= sin^{-1}(sin 3y)

= 3y

= 3 sin^{-1} x

= LHS

Question 2:

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

Answer:

Let cos^{-1} x = y, then cos y = x

RHS:

cos^{-1}(4x^{3} – 3x) = cos^{-1}(4cos^{3} 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 + x^{2}) - 1}/x, x ≠ 0, in the simplest form.

Answer:

Let x = tan y

Now, tan^{-1}{√(1 + x^{2}) - 1}/x = tan^{-1}[{√(1 + tan^{2} y) - 1}/tan y]

= tan^{-1}[(sec y – 1)/tan y]

= tan^{-1}[(1 – cos y)/sin y]

= tan^{-1}[(2sin^{2} 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/√(x^{2} – 1)}, |x| > 1, in the simplest form.

Answer:

Let x = cosec y

Now, tan^{-1}{1/√(1 + x^{2})} = tan^{-1}{1/√(cosec^{2} 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}[√{(2sin^{2} x/2)/ (2cos^{2} x/2)}]

= tan^{-1}[√(tan^{2} 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/√(a^{2} – x^{2})], |x| < a in the simplest form.

Answer:

Let x = a sin y

Now, tan^{-1}[x/√(a^{2} – x^{2})] = tan^{-1}[a sin y/√(a^{2} – a^{2} sin^{2} y)]

= tan^{-1}[a sin y/{a√ (1 – sin^{2} 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}[(3a^{2}x – x^{3})/(a^{3} – 3ax^{2})], a > 0, -a/√3 ≤ x ≤ a/√3 the simplest form.

Answer:

Let x = a tan y

Now, tan^{-1}[(3a^{2}x – x^{3})/(a^{3} – 3ax^{2})] = tan^{-1}[(3a^{2} * a tan y – a^{3} tan^{3} y)/(a^{3} – 3a * a^{2} tan^{2} y)]

= tan^{-1}[(3a^{3} tan y – a^{3} tan^{3} y)/(a^{3} – 3a^{3} tan^{2} y)]

= tan^{-1}[(3 tan y – tan^{3} y)/(1 – 3 tan^{2} 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 + x^{2})} + cos^{-1}{(1 – y^{2})/(1 + y^{2})}]/2, |x| < 1, y > 0 and xy < 1

Answer:

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

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

=> (2x^{2} – 4)/(-3) = 1

=> 2x^{2} – 4 = -3

=> 2x^{2} = -3 + 4

=> 2x^{2} = 1

=> x^{2} = 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/√(5^{2} - 3^{2})} + 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/√(5^{2} - 3^{2})}

= 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/√(17^{2} - 8^{2})} + tan^{-1} {3/√(5^{2} - 3^{2})}

= 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} {√(5^{2} - 4^{2})/4} + tan^{-1} {√(13^{2} - 12^{2})/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/√(56^{2} + 33^{2})}

= 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} {√(13^{2} - 12^{2})/12} + tan^{-1} {3/√(5^{2} - 3^{2})}

= 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/√(56^{2} + 33^{2})}

= 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/√(13^{2} - 5^{2})} + tan^{-1} {√(5^{2} - 3^{2})/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 – x^{2})/(1 + x^{2})}]

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

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

= (1 + cos x)/sin x

= (2 * cos^{2} 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} [{√(2cos^{2} y/2) - √(2sin^{2} y/2)}/{√(2cos^{2} y/2) + √(2sin^{2} 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} √(3^{2} – 1^{2})/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 – cos^{2} x)} = tan^{-1}(2cosec x) [Since 2tan^{-1} x = tan^{-1}{2x/(1 – x^{2})}]

= 2cos x/(1 – cos^{2} x) = 2cosec x

= 2cos x/sin^{2} x = 2/sin x

= sin x * cos x = sin^{2} x

= sin x * cos x - sin^{2} 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 – x^{2}) (B) 1/√(1 – x^{2}) (C) 1/√(1 + x^{2}) (D) x/√(1 + x^{2})

Answer:

Given, sin(tan^{-1} x) = sin(sin^{-1}{x/√(1 + x^{2})}) [Since tan^{-1} a/b = sin^{-1}{a/√(a^{2} + b^{2})}]

= x/√(1 + x^{2})

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 – 2sin^{2} y

=> 2sin^{2} y – sin y = 0

=> 2x^{2} – 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} [{x^{2} + xy - xy + y^{2}}/{xy + y^{2} + x^{2} – xy}]

= tan^{-1} [{x^{2} + y^{2}}/{y^{2} + x^{2}}]

= tan^{-1} 1

= π/4

Hence, the correct answer is option (C).

.