RD Sharma Class 12 Exercise 3.6 Inverse Trigonometric Functions Solutions Maths

RD Sharma Class 12 Exercise 3.6 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 12:24 PM IST

You can find free RD Sharma class 12 chapter 3 Exercise 3.6 solution here. For Inverse Trigonometric Functions, all RD Sharma Book Solutions are listed here, exercise by exercise. RD Sharma Solutions can assist you with preparing for a variety of competitive exams in high school, graduate school, and undergraduate education. In addition, it has been proven that practicing questions with RD Sharma Class 12th Exercise 3.6 Solution improves math skills.

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RD Sharma Class 12 Solutions Chapter 3 Inverse Trigonometric Functions - Other Exercise

Inverse Trigonometric Functions Excercise: 3.6

Inverse Trigonometric Function Exercise 3.6 Question 1 (i).

Answer:

\frac{5 π}{6}

Hints: The

\cot^{- 1}

function is defined as a function whose domain is R and the principal value branch of the function

c o t^{- 1}

(0, π)

Thus,

c o t^{- 1} : R \to (0, π)

Given:

c o t^{- 1} (\sqrt{- 3})

Solution:
Let

y = c o t^{- 1} (\sqrt{- 3}) \cdot \cdot \cdot (1)

c o t y = - \sqrt{3}

c o t y = - c o t \frac{π}{6} [c o t \frac{π}{6} = \sqrt{3}]

c o t y = c o t (π - \frac{π}{6}) [∵ c o t (π - θ) = - c o t θ]

c o t y = c o t \frac{5 π}{6}

y = \frac{5 π}{6}

c o t^{- 1} (\sqrt{- 3}) = \frac{5 π}{6}

(From equation 1)

∵

The principal value branch of the function

c o t^{- 1} i s (0, π)

c o t^{- 1} (\sqrt{- 3}) = \frac{5 π}{6} ϵ (0, π)

Hence the principal value of

c o t^{- 1} (\sqrt{- 3})

\frac{5 π}{6}

Inverse Trigonometric Function Exercise 3.6 Question 1 (ii).

Answer:

\frac{π}{6}

Hint: The

c o t^{- 1}

function is defined as a function whose domain R and the principal value branch of the function

c o t^{- 1}

(0, π)

. Thus

c o t^{- 1} : R \to (0, π)

Given:

c o t^{- 1} (\sqrt{3})

Solution:
Let

y = c o t^{- 1} (\sqrt{3})

\cdot \cdot \cdot 1

c o t y = \sqrt{3}

c o t y = c o t \frac{π}{6} [c o t \frac{π}{6} = \sqrt{3}]

y = \frac{π}{6}

c o t^{- 1} \sqrt{3} = \frac{π}{6}

(From equation 1)

∵

The principal value branch of the function

c o t^{- 1}

(0, π)

c o t^{- 1} (\sqrt{3})

= \frac{π}{6} ϵ (0, π)

Hence the principal value of

c o t^{- 1} (\sqrt{3})

\frac{π}{6}

Inverse Trigonometric Function Exercise 3.6 Question 1 (iii).

Answer:

\frac{2 π}{3}

Hint: The

c o t^{- 1}

function is defined as a function whose domain R and the principal value branch of the function

c o t^{- 1}

(0, π)

. Thus

c o t^{- 1} : R \to (0, π)

Given:

c o t^{- 1} (\frac{- 1}{\sqrt{3}})

Solution:
Let

y = c o t^{- 1} (\frac{- 1}{\sqrt{3}})

\cdot \cdot \cdot 1

c o t y = \frac{- 1}{\sqrt{3}}

c o t y = - c o t \frac{π}{3} [c o t \frac{π}{3} = \frac{1}{\sqrt{3}}]

c o t y = c o t (π - \frac{π}{3}) [∵ c o t (π - θ) = - c o t θ]

y = \frac{2 π}{3}

c o t^{- 1} (\frac{- 1}{\sqrt{3}}) = \frac{2 π}{3}

(From equation 1)

∵

The principal value branch of the function

c o t^{- 1}

(0, π)

c o t^{- 1} (\frac{- 1}{\sqrt{3}})

= \frac{2 π}{3} ϵ (0, π)

Hence the principal value of

c o t^{- 1} (\frac{- 1}{\sqrt{3}})

\frac{2 π}{3}

Inverse Trigonometric Function Exercise 3.6 Question 1 (iv).

Answer: $\frac{3 π}{4}$
Hint: The $\cot^{- 1}$ function is defined as a function whose domain R and the principal value branch of the function $\cot^{- 1}$ is $(0, π) .$ Thus $c o t^{- 1} : R \to (0, π)$
First we will convert $t a n \frac{3 π}{4}$ into $\cot$
Given: $c o t^{- 1} (\tan \frac{3 π}{4})$
Solution:
Let $y = c o t^{- 1} (\tan \frac{3 π}{4})$ $\cdot \cdot \cdot 1$
$c o t y = \tan \frac{3 π}{4}$
$c o t y = c o t (\frac{π}{2} - \frac{3 π}{2}) [∵ t a n θ = \cot (\frac{π}{2} - θ)]$
$c o t y = c o t (\frac{- π}{4})$
$c o t y = - c o t (\frac{π}{4}) [∵ c o t (- θ) = - c o t θ]$
$c o t y = c o t (π - \frac{π}{4}) [c o t (π - θ) = - c o t θ]$
$c o t y = c o t (\frac{3 π}{4})$
$y = \frac{3 π}{4}$
$c o t^{- 1} (t a n \frac{3 π}{4}) = \frac{3 π}{4}$ (From equation 1)
$∵$ The principal value branch of the function $c o t^{- 1}$ is $(0, π)$
$c o t^{- 1} (t a n \frac{3 π}{4}) = \frac{3 π}{4} ϵ (0, π)$
Hence the principal value of $c o t^{- 1} (t a n \frac{3 π}{4})$ is $\frac{3 π}{4}$

Inverse Trigonometric Function Exercise 3.6 Question 2.

Answer:

R - {n π, n ϵ Z}

Hints: First we will write the domain of

\cot x

and

\cot^{- 1} x

respectively. After that we will solve for common domain
Given:

f (x) = \cot x + \cot^{- 1} (x)

Solution:
Let us first solve for

c o t x

∵

\cot x

is defined for

x ϵ R - {n π, n ϵ z}

∵

Domain of

\cot x = R - {n π, n ϵ z}

\cdot \cdot \cdot \cdot 1

Let us solve for

\cot^{- 1} (x)

∵

\cot^{- 1} x

is defined for

x ϵ R

∵

Domain of

\cot^{- 1} (x) = R

\cdot \cdot \cdot \cdot 2

Now to find the domain of

f (x)

we take the intersection of both equations 1 and 2
(Domain of

\cot x

)

\cap

(Domain of

\cot^{- 1} x

)

= R - {n π, n ϵ z}

Hence the domain of

f (x) = \cot x + \cot^{- 1} x

R - {n π, n ϵ z}

Inverse Trigonometric Function Exercise 3.6 Question 3 (i).

Answer:

\frac{2 π}{3}

Hints:
The principal value branch of the function

\cot^{- 1} i s (0, π)

The principal value branch of the function

\sec^{- 1} i s [0, π] - {\frac{π}{2}}

The principal value branch of the function cosec -1 is

[\frac{- π}{2}, \frac{π}{2}] - {0}

We will find all the principal values between these intervals
Given:

\cot^{- 1} (\frac{1}{\sqrt{3}}) - c o s e c^{- 1} (- 2) + \sec^{- 1} (\frac{2}{\sqrt{3}})

Solution:

\cot^{- 1} (\frac{1}{\sqrt{3}}) - c o s e c^{- 1} (- 2) + \sec^{- 1} (\frac{2}{\sqrt{3}})

\cdot \cdot \cdot \cdot (i)

Let us first solve for

\cot^{- 1} (\frac{1}{\sqrt{3}})

Let

x = c o t^{- 1} (\frac{1}{\sqrt{3}})

\cdot \cdot \cdot \cdot (i i)

\cot x = \frac{1}{\sqrt{3}}

\cot x = \cot (\frac{π}{3}) [\cot \frac{π}{3} = \frac{1}{\sqrt{3}}]

x = \frac{π}{3}

\cot^{- 1} (\frac{1}{\sqrt{3}}) = \frac{π}{3}

{from equation (ii)}
The principal value branch of

\cot^{- 1} i s (0, π)

∵

\cot^{- 1} (\frac{1}{\sqrt{3}}) = \frac{π}{3} ϵ (0, π)

Therefore principal value of

\cot^{- 1} (\frac{1}{\sqrt{3}})

\frac{π}{3}

\cdot \cdot \cdot (i i i)

Let us solve for

c o s e c^{- 1} (- 2)

Let

y = c o s e c^{- 1} (- 2)

\cdot \cdot \cdot (i v)

c o s e c y = - 2

c o s e c y = - c o s e c \frac{π}{6} [c o s e c \frac{π}{6} = 2]

c o s e c y = c o s e c (\frac{- π}{6})

y = - \frac{π}{6}

c o s e c^{- 1} (- 2) = - \frac{π}{6}

{from equation (iv)}
The principal value branch of

c o s e c^{- 1} i s [\frac{- π}{2}, \frac{π}{2}] - {0}

∵ c o s e c^{- 1} (- 2) = - \frac{π}{6} ϵ [\frac{- π}{2}, \frac{π}{2}] - {0}

Therefore principal value of

c o s e c^{- 1} (- 2) i s \frac{- π}{6}

… (v)
Let us solve for

\sec^{- 1} (\frac{2}{\sqrt{3}})

Let

z = \sec^{- 1} (\frac{2}{\sqrt{3}}) \cdot \cdot \cdot (v i)

\begin{aligned} \sec z = \frac{2}{\sqrt{3}} [\sec (\frac{π}{6}) = \frac{2}{\sqrt{3}}] \\ \sec z = \sec \frac{π}{6} \\ z = \frac{π}{6} \\ \sec^{- 1} (\frac{2}{\sqrt{3}}) = \frac{π}{6} \end{aligned}

{from equation (vi)}
The principal value branch of the function

\sec^{- 1}

[0, π] - {\frac{π}{2}}

∵ \sec^{- 1} (\frac{2}{\sqrt{3}}) = \frac{π}{6} ϵ [0, π] - {\frac{π}{2}}

Therefore the principal value of

\sec^{- 1} (\frac{2}{\sqrt{3}}) i s \frac{π}{6}

… (vii)
Now from equation (i) –

\cot^{- 1} (\frac{1}{\sqrt{3}}) - c o s e c^{- 1} (- 2) + \sec^{- 1} (\frac{2}{\sqrt{3}})

Putting the value of

\cot^{- 1} (\frac{1}{\sqrt{3}}), c o s e c^{- 1} (- 2) a n d \sec^{- 1} (\frac{2}{\sqrt{3}})

from equations (iii), (v) and (vii) respectively.

\begin{aligned} = \frac{π}{3} - (- \frac{π}{6}) + \frac{π}{6} \\ = \frac{π}{3} + (\frac{π}{6}) + \frac{π}{6} \\ = \frac{2 π + π + π}{6} \end{aligned}

= \frac{4 π}{6} = \frac{2 π}{3}

Inverse Trigonometric Function Exercise 3.6 Question 3 (ii).

Answer:

\frac{π}{4}

Hints:
First we will find the principal value of

\sin^{- 1} \frac{\sqrt{3}}{2}

The principal value branch of function

\sin^{- 1} i s [\frac{- π}{2}, \frac{π}{2}]

The principal value branch of function

\cot^{- 1} i s (0, π)

Given:

c o t^{- 1} {2 \cos (\sin^{- 1} \frac{\sqrt{3}}{2})}

Solution:

c o t^{- 1} {2 \cos (\sin^{- 1} \frac{\sqrt{3}}{2})}

\cdot \cdot \cdot (i)

Let us first solve for

\sin^{- 1} \frac{\sqrt{3}}{2}

Let

x = \sin^{- 1} \frac{\sqrt{3}}{2}

\cdot \cdot \cdot (i i)

\sin x = \frac{\sqrt{3}}{2}

\sin x = \sin \frac{π}{3}

x = \frac{π}{3}

\sin^{- 1} \frac{\sqrt{3}}{2} = \frac{π}{3}

[from equation (ii)]
The principal value branch of

\sin^{- 1}

[\frac{- π}{2}, \frac{π}{2}]

∵ \sin^{- 1} \frac{\sqrt{3}}{2} = \frac{π}{3} ϵ [\frac{- π}{2}, \frac{π}{2}]

∵

Principal value of

\sin^{- 1} \frac{\sqrt{3}}{2}

\frac{π}{3}

\cdot \cdot \cdot (i i i)

Now from equation (i)

c o t^{- 1} {2 \cos (\sin^{- 1} \frac{\sqrt{3}}{2})}

= \cot^{- 1} {2 \cos (\frac{π}{3})}

[from equation (iii)]

= \cot^{- 1} (2 \times \frac{1}{2})

= c o t^{- 1} (1)

\cdot \cdot \cdot (i v)

Let

y = c o t^{- 1} (1)

\cdot \cdot \cdot (v)

\cot y = 1

\cot y = \cot \frac{π}{4}

y = \frac{π}{4}

[from v]

\cot^{- 1} (1) = \frac{π}{4} ϵ (0, π)

From equation (iv)

c o t^{- 1} {2 \cos (\sin^{- 1} \frac{\sqrt{3}}{2})}

= \cot^{- 1} (1)

= \frac{π}{4}

Inverse Trigonometric Function Exercise 3.6 Question 3 (iii).

Answer:

\frac{π}{6}

Hints:
The principal value branch of the function

c o s e c^{- 1} i s [\frac{- π}{2}, \frac{π}{2}] - {0}

The principal value branch of the function

\cot^{- 1} i s (0, π)

Given:

c o s e c^{- 1} (\frac{- 2}{\sqrt{3}}) + 2 \cot^{- 1} (- 1)

Solution:

c o s e c^{- 1} (\frac{- 2}{\sqrt{3}}) + 2 \cot^{- 1} (- 1)

\cdot \cdot \cdot (i)

Let us first solve for

c o s e c^{- 1} (\frac{- 2}{\sqrt{3}})

Let

x = c o s e c^{- 1} (\frac{- 2}{\sqrt{3}})

\cdot \cdot \cdot (i i)

c o s e c x = - \frac{2}{\sqrt{3}}

c o s e c x = - c o s e c \frac{π}{3}

c o s e c x = c o s e c (\frac{- π}{3})

x = - \frac{π}{3}

c o s e c^{- 1} (\frac{- 2}{\sqrt{3}}) = - \frac{π}{3}

[from equation (ii)]
The principal value branch of the function

\sec^{- 1} i s [0, π] - {\frac{π}{2}}

∵ \sec^{- 1} (\frac{2}{\sqrt{3}}) = \frac{- π}{3} ϵ [0, π] - {\frac{π}{2}}

Therefore the principal value of

\sec^{- 1} (\frac{2}{\sqrt{3}})

\frac{- π}{3}

\cdot \cdot \cdot (i i i)

Now let us solve for

\cot^{- 1} (- 1)

Let

y = \cot^{- 1} (- 1) \cdot \cdot \cdot (i v)

\cot y = - 1 [c o t (\frac{π}{4}) = 1]

\cot y = - \cot \frac{π}{4}

\cot y = \cot (π - \frac{π}{4}) [\cot (π - θ) = - c o t θ]

\cot y = \cot (\frac{3 π}{4})

y = \frac{3 π}{4}

\cot^{- 1} (- 1) = \frac{3 π}{4}

[from equation (iv) ]
The principal value branch of the

\cot^{- 1}

(0, π)

∵ \cot^{- 1} (- 1) = \frac{3 π}{4} ϵ (0, π)

∵

Principal value of

\cot^{- 1} (- 1) = \frac{3 π}{4} \cdot \cdot \cdot (v)

Now, from equation (i)

c o s e c^{- 1} (\frac{- 2}{\sqrt{3}}) + 2 \cot^{- 1} (- 1)

From equation (iii) and (v)

= - \frac{π}{3} + 2 \times \frac{3 π}{4}

= - \frac{π}{3} + \frac{3 π}{2}

= \frac{- 2 π + 9 π}{6}

= \frac{7 π}{6}

Inverse Trigonometric Function Exercise 3.6 Question 3 (iv).

Answer: $\frac{- π}{12}$
Hints: First we will find principal values of $\tan^{- 1} (\frac{- 1}{\sqrt{3}}), \cot^{- 1} (\frac{1}{\sqrt{3}}) a n d \tan^{- 1} (\sin (\frac{- π}{2}))$
Given: $\tan^{- 1} (\frac{- 1}{\sqrt{3}}) + \cot^{- 1} (\frac{1}{\sqrt{3}}) + \tan^{- 1} (\sin (\frac{- π}{2}))$
Solution:
$\tan^{- 1} (\frac{- 1}{\sqrt{3}}) + \cot^{- 1} (\frac{1}{\sqrt{3}}) + \tan^{- 1} (\sin (\frac{- π}{2}))$ $\cdot \cdot \cdot \cdot (i)$
Let $x = \tan^{- 1} (\frac{- 1}{\sqrt{3}}) \cdot \cdot \cdot \cdot (i i)$
$\begin{aligned} \tan x = - \frac{1}{\sqrt{3}} \\ \tan x = - \tan \frac{π}{6} \\ \tan x = \tan (\frac{- π}{6}) \\ x = \frac{- π}{6} \end{aligned}$
$\tan^{- 1} (\frac{- 1}{\sqrt{3}}) = \frac{- π}{6}$ [from equation]
The range of principal value branch of $\tan^{- 1}$ is $[\frac{- π}{2}, \frac{π}{2}]$
$∵ \tan^{- 1} (\frac{- 1}{\sqrt{3}}) = \frac{- π}{6} ϵ [\frac{- π}{2}, \frac{π}{2}]$
Hence principal value of $\tan^{- 1} (\frac{- 1}{\sqrt{3}}) = \frac{- π}{6} \cdot \cdot \cdot \cdot (i i i)$
Now let us solve for $\tan^{- 1} (- \frac{1}{\sqrt{3}})$
Let $y = \cot^{- 1} (\frac{1}{\sqrt{3}}) \cdot \cdot \cdot (i v)$
$\cot x = \frac{1}{\sqrt{3}}$
$\cot x = \cot (\frac{π}{3}) [\cot \frac{π}{3} = \frac{1}{\sqrt{3}}]$
$x = \frac{π}{3}$
$\cot^{- 1} (\frac{1}{\sqrt{3}}) = \frac{π}{3}$ {from equation (iv)}
The principal value branch of $\cot^{- 1}$ is $(0, π)$
$∵ \cot^{- 1} (\frac{1}{\sqrt{3}}) = \frac{π}{3} ϵ (0, π)$
Therefore principal value of $\cot^{- 1} (\frac{1}{\sqrt{3}}) i s \frac{π}{3}$
Now let us solve for $\tan^{- 1} (\sin (\frac{- π}{2}))$
Let $z = \tan^{- 1} (\sin (\frac{- π}{2})) \cdot \cdot \cdot (v i)$
$\tan z = \sin (\frac{- π}{2})$
$\tan z = - s i n (\frac{π}{2}) [\sin (\frac{- π}{2}) = 1]$
$\begin{aligned} \begin{matrix} \tan z = - 1 \\ \tan z = - \tan \frac{π}{4} \end{matrix} \\ \tan z = \tan \frac{- π}{4} \\ z = \frac{- π}{4} \end{aligned}$
$\tan^{- 1} (\sin (\frac{- π}{2})) = \frac{- π}{4}$
The range of principal value branch of $\tan^{- 1} i s [\frac{- π}{2}, \frac{π}{2}]$
$∵ \tan^{- 1} (\sin (\frac{- π}{2})) = \frac{- π}{4} ϵ [\frac{- π}{2}, \frac{π}{2}]$
$∵$ The principal value of $\tan^{- 1} (\sin (\frac{- π}{2})) i s \frac{- π}{4} \cdot \cdot \cdot \cdot \cdot (v i i)$
Now from equation (i)
$\tan^{- 1} (\frac{- 1}{\sqrt{3}}) + \cot^{- 1} (\frac{1}{\sqrt{3}}) + \tan^{- 1} (\sin (\frac{- π}{2}))$
Putting the value of $\tan^{- 1} (\frac{- 1}{\sqrt{3}}), \cot^{- 1} (\frac{1}{\sqrt{3}}) a n d \tan^{- 1} (\sin (\frac{- π}{2}))$ from equations (iii), (v) and (vii) respectively.
$\begin{aligned} = \frac{- π}{6} + \frac{π}{3} + (\frac{- π}{4}) \\ = \frac{- π}{6} + \frac{π}{3} - \frac{π}{4} \\ = \frac{- 2 π + 4 π - 3 π}{12} \end{aligned}$
$= \frac{- π}{12}$

Class 12 RD Sharma Chapter 3 Exercise 3.6 solution has nine questions, including subparts, which cover questions regarding the principal value, domain, and range of inverse trigonometric functions, as well as several key qualities that can aid you in solving inverse trigonometric function problems. RD Sharma Class 12th Exercise 3.6, inverse trigonometric functions can be used to calculate the angle from any trigonometric ratio. All of these proportions have applications in various domains, including engineering, geometry, and physics.

Experts in the field create answers based on the students' comprehension abilities. Thus, it aids students in honing their logical reasoning abilities, which are critical for exam success. The inverse of the cotangent function is the sole topic of this practice. RD Sharma Class 12 solution Inverse Trigonometric Function Ex 3.6 PDF can be used as a study aid by students who desire to do well on the exam.

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