RD Sharma Class 12 Exercise 3.13 Inverse Trigonometric Functions Solutions Maths

RD Sharma Class 12 Exercise 3.13 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 11:46 AM IST

The RD Sharma solution books are the best companion for the class 12 students. When a teacher is the best guide at the school to clear the doubts, the RD Sharma solution books are the best guides for the students at their home. Many students find it challenging to cope with the concepts in the 3rd chapter of mathematics, Inverse Trigonometry. The RD Sharma Class 12th Exercise 3.13 books are a boon for them.

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Inverse Trigonometry is not an easy chapter to solve, and there are numerous formulas and methods to bring the right solutions. Especially in the thirteenth exercise, it is even more challenging for students to solve when they are unclear about the previous exercises. Exercise 3.13 consists of problems that include the concepts like sine, cosine, tangent, cotangent, secant, and cosecant functions. There are five questions, including its subparts, to be solved in this exercise. To make it easier for the students, the RD Sharma Class 12 Chapter 3 Exercise 3.13 will lend a helping hand to the students.

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

Inverse Trigonometry Function Ex 3.13

Inverse Trigonometric Function Exercise 3.13 Question 1

Answer:

Given:

\cos^{- 1} \frac{x}{2} + \cos^{- 1} \frac{y}{3} = α

To prove:

9 x^{2} - 12 x y \cos α + 4 y^{2} = 36 \sin^{2} α

Hint:
We are applying the formula of

\cos^{- 1} x + \cos^{- 1} y = \cos^{- 1} [x y - (\sqrt{1 - x^{2}}) (\sqrt{1 - y^{2}})]

Solution:
We have

\cos^{- 1} \frac{x}{2} + \cos^{- 1} \frac{y}{3} = \cos^{- 1} [\frac{x}{2} \times \frac{y}{3} - (\sqrt{1 - {(\frac{x}{2})}^{2}}) (\sqrt{1 - {(\frac{y}{3})}^{2}})]

= \cos^{- 1} [\frac{x y}{6} - \frac{\sqrt{4 - x^{2}}}{2} \times \frac{\sqrt{9 - y^{2}}}{3}]

= \cos^{- 1} [\frac{x y - \sqrt{4 - x^{2}} \times \sqrt{9 - y^{2}}}{6}] = α

(Let)

\Rightarrow x y - \sqrt{4 - x^{2}} \times \sqrt{9 - y^{2}} = 6 \cos α

\Rightarrow (x y - 6 \cos α) = \sqrt{4 - x^{2}} \times \sqrt{9 - y^{2}}

On squaring both sides, we get

\Rightarrow {(x y - 6 \cos α)}^{2} = (4 - x^{2}) (9 - y^{2}) \Rightarrow x^{2 y^{2}} + 36 \cos^{2} α - 12 x y \cos α = 36 - 9 x^{2} - 4 y^{2} + x^{2} y^{2}

\Rightarrow 9 x^{2} + 4 y^{2} - 36 + 36 \cos^{2} α - 12 x y \cos α = 0 \Rightarrow 9 x^{2} + 4 y^{2} - 12 x y \cos α - 36 (1 - \cos^{2} α) = 0

\Rightarrow 9 x^{2} + 4 y^{2} - 12 x y \cos α - 36 \sin^{2} α = 0

\Rightarrow 9 x^{2} + 4 y^{2} - 12 x y \cos α = 36 \sin^{2} α

Hence proved

Inverse Trigonometric Function Exercise 3.13 Question 2

Answer:

x = a b

Given:

\cos^{- 1} \frac{a}{x} - \cos^{- 1} \frac{b}{x} = \cos^{- 1} \frac{1}{b} - \cos^{- 1} \frac{1}{a}

Hint:
First, we will separate the angle of

a

terms then we will apply the formula,

\cos^{- 1} x + \cos^{- 1} y = \cos^{- 1} [x y - (\sqrt{1 - x^{2}}) (\sqrt{1 - y^{2}})]

\cos^{- 1} \frac{a}{x} + \cos^{- 1} \frac{1}{a} = \cos^{- 1} \frac{1}{b} + \cos^{- 1} \frac{b}{x}

Solution:
We know that,

\cos^{- 1} x + \cos^{- 1} y = \cos^{- 1} [x y - (\sqrt{1 - x^{2}}) (\sqrt{1 - y^{2}})]

Substituting the values in the formula, we get

\cos^{- 1} [\frac{1}{x} - \sqrt{1 - {(\frac{a}{x})}^{2}} \sqrt{1 - {(\frac{1}{a})}^{2}}] = \cos^{- 1} [\frac{1}{x} \sqrt{1 - {(\frac{b}{x})}^{2}} \sqrt{1 - {(\frac{1}{b})}^{2}}]

\Rightarrow \frac{1}{x} - \sqrt{1 - {(\frac{a}{x})}^{2}} \sqrt{1 - {(\frac{1}{a})}^{2}} = \frac{1}{x} - \sqrt{1 - {(\frac{b}{x})}^{2}} \sqrt{1 - {(\frac{1}{b})}^{2}}

\Rightarrow \sqrt{1 - {(\frac{a}{x})}^{2}} \sqrt{1 - {(\frac{1}{a})}^{2}} = \sqrt{1 - {(\frac{b}{x})}^{2}} \sqrt{1 - {(\frac{1}{b})}^{2}}

Squaring on both sides, we get

\Rightarrow (1 - {(\frac{a}{x})}^{2}) (1 - {(\frac{1}{a})}^{2}) = (1 - {(\frac{b}{x})}^{2}) (1 - {(\frac{1}{b})}^{2})

\Rightarrow 1 - {(\frac{a}{x})}^{2} - {(\frac{1}{a})}^{2} + {(\frac{1}{x})}^{2} = 1 - {(\frac{b}{x})}^{2} - {(\frac{1}{b})}^{2} + {(\frac{1}{x})}^{2}

\Rightarrow {(\frac{b}{x})}^{2} - {(\frac{a}{x})}^{2} = {(\frac{1}{a})}^{2} - {(\frac{1}{b})}^{2}

On Simplifying, we get

\Rightarrow (b^{2} - a^{2}) a^{2} b^{2} = x^{2} (b^{2} - a^{2})

\Rightarrow x^{2} = a^{2} b^{2}

\Rightarrow x = a b

Hence the result.

Inverse Trigonometric Function Exercise 3.13 Question 3

Answer:

x = \frac{1}{2}

Given:

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

Hint:
We are applying the formula,

\cos^{- 1} x + \cos^{- 1} y = \cos^{- 1} [x y - (\sqrt{1 - x^{2}}) (\sqrt{1 - y^{2}})]

Solution:
We have

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

Using formula written on the hint, we get

\Rightarrow \cos^{- 1} [\sqrt{3 x} \times x - (\sqrt{1 - {(\sqrt{3 x})}^{2}}) (\sqrt{1 - x^{2}})] = \frac{x}{2}

\Rightarrow \cos^{- 1} [\sqrt{3 x^{2}} - (\sqrt{1 - 3 x^{2}}) (\sqrt{1 - x^{2}})] = \frac{x}{2}

\Rightarrow \sqrt{3 x^{2}} - (\sqrt{1 - 3 x^{2}}) (\sqrt{1 - x^{2}}) = \cos \frac{x}{2}

\Rightarrow \sqrt{3 x^{2}} - (\sqrt{1 - 3 x^{2}}) (\sqrt{1 - x^{2}}) = 0 [∵ \cos \frac{x}{2} = 0]

\Rightarrow \sqrt{3 x^{2}} - (\sqrt{1 - 3 x^{2}}) (\sqrt{1 - x^{2}})

Squaring on both sides, we get

\Rightarrow 3 x^{4} - (1 - 3 x^{2}) (1 - x^{2})

\Rightarrow 3 x^{4} = 1 - x^{2} - 3 x^{2} + 3 x^{4}

\Rightarrow 1 - 4 x^{2} = 0

\Rightarrow 4 x^{2} = 1

\Rightarrow x^{2} = \frac{1}{4}

\Rightarrow x = \pm \frac{1}{2} [A s - \frac{1}{2} i s n o t s a t i s f y i n g t h e e q u a t i o n]

Hence the result is

x = \frac{1}{2}

Inverse Trigonometric Function Exercise 3.13 Question 4

Answer:

Given:

\cos^{- 1} (\frac{4}{5}) + \cos^{- 1} (\frac{12}{13})

To prove:

\cos^{- 1} (\frac{4}{5}) + \cos^{- 1} (\frac{12}{13}) = \cos^{- 1} (\frac{33}{65})

Hint:
We will use the formula on L.H.S

\cos^{- 1} x + \cos^{- 1} y = \cos^{- 1} [x y - (\sqrt{1 - x^{2}}) (\sqrt{1 - y^{2}})]

Solution:
Taking L.H.S
L.H.S

\cos^{- 1} (\frac{4}{5}) + \cos^{- 1} (\frac{12}{13})

\Rightarrow \cos^{- 1} [\frac{48}{65} - \sqrt{1 - \frac{16}{25}} \sqrt{1 - \frac{144}{169}}]

\Rightarrow \cos^{- 1} [\frac{48}{65} - \frac{3}{5} \times \frac{5}{13}]

\Rightarrow \cos^{- 1} (\frac{33}{65})

=

R.H.S
Hence we get R.H.S.

Inverse Trigonometric Function Exercise 3.13 Question 5

Answer:

Given:

\cos^{- 1} \frac{12}{13} + \sin^{- 1} \frac{3}{5}

Hint:
First, we convert

\cos^{- 1}

\sin^{- 1}

and then use

\sin (a + b)

formula.
Solution:
Let

a = \cos^{- 1} \frac{12}{13}

and

b = \sin^{- 1} \frac{3}{5}

Finding

\sin a, \cos a

.
Let

a = \cos^{- 1} \frac{12}{13}

\cos a = (\frac{12}{13})

We know that

\sin a = \sqrt{1 - \cos^{2}} a

= \sqrt{1 - {(\frac{12}{13})}^{2}}

= \sqrt{\frac{25}{169}}

= \frac{5}{13}

Again, finding

\sin b, \cos b

Let

b = \sin^{- 1} (\frac{3}{5})

\Rightarrow \sin b = \frac{3}{5}

We know that,

\cos b = \sqrt{1 - \sin^{2} b}

\Rightarrow \cos b = \sqrt{1 - {(\frac{3}{5})}^{2}}

= \sqrt{\frac{16}{25}} = \frac{4}{5}

\cos b = \frac{4}{5}

We know that,

\sin (a + b) = \sin a \cos b + \cos a \sin b

Putting,

\sin a = \frac{5}{13}, \cos a = \frac{12}{13}

\sin b = \frac{3}{5}, \cos b = \frac{4}{5}

We have,

\sin (a + b) = \frac{5}{13} \times \frac{4}{5} + \frac{12}{13} \times \frac{3}{5}

= \frac{20}{65} + \frac{36}{65}

\sin (a + b) = \frac{56}{65}

a + b = \sin^{- 1} \frac{56}{65}

Putting the value of

a, b

\cos^{- 1} \frac{12}{13} + \sin^{- 1} \frac{3}{5} = \sin^{- 1} \frac{56}{65}

LHS=RHS
Hence proved.

The hesitations regarding the accuracy of the answers need not prevail. Every solution is provided by the best team of teachers and professors who are experts in the educational field. They have also followed the NCERT pattern to make it easy for the CBSE board students to follow. You will find the Inverse Trigonometry chapter very easy to solve once you start using the Class 12 RD Sharma Chapter 3 Exercise 3.13 Solution book.

This material gives you the freedom to choose the method you want to adapt in finding answers. This is due to the presence of elaborated solutions as well as shortcuts techniques given in the book. Moreover, the students can download the RD Sharma Class 12 Solutions Inverse Trigonometry Ex 3.13 from top educational websites like Career 360. And the bonus surprise is that you need not pay even a single rupee to access this solution book. Everything is available free of cost for the benefit of the students.

The wide usage of the RD Sharma solutions book and its best solutions has led to its strong recognition. Hence, there are many chances that your class 12 public mathematics exam will have questions from this book. Therefore, the RD Sharma Class 12 Solutions Chapter 3 ex 3.13 material is suitable for your exam preparations. Make it a practice by using this book for solving your homework and assignments. Download your copy of the RD Sharma solutions book from the Career360 website and start to solve the sums accordingly.

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Frequently Asked Questions (FAQs)

1. What book should I refer to to clear my doubts on Chapter 3, Inverse Trigonometry?

The RD Sharma Class 12th exercise 3.13 solutions book is enough to solve all your doubts regarding Inverse Trigonometry.

2. Which is the best guide for class 12 students to prepare well for their public exam?

Experts in the educational sector provide the solutions in the RD Sharma books. The students can use it to prepare for their exams, homework, and assignments.

3. Can every student refer to the RD Sharma solutions book?

rrespective of how they perform in exams, every student can use the RD Sharma solutions book to learn the concepts. The solutions are given in various methods; hence the students can prefer the best way they find easy and go ahead with it.

4. Is it enough to work out the sums for chapter 3 given in the RD Sharma book?

The RD Sharma Class 12 Chapter 3 Exercise 3.13 solutions are the best material. As there are many practice questions, students can attain the speed to solve sums effectively with a bit of practice.

5. How much does the RD Sharma Class 12th exercise 3.13 solutions book cost?

You can download the best set of RD Sharma Class 12th exercise 3.13 solutions from the Career360 website for free of cost. So be it, students or teachers, no one is asked to pay money for this material. Everyone can use this resource for free.

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RD Sharma Class 12 Exercise 3.13 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

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

Inverse Trigonometry Function Ex 3.13

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