# Plus Two Math's Solution Ex 2.2 Chapter2 Inverse Trigonometric Functions

NCERT Solutions PDF for Plus Two Mathematics Chapter-2 is available for download. You can find the official PDF on the Keralanotes website and download it from there. Reverse trigonometry function plus two complete PDF solutions plus 2 for Chapter 2 of NCERT Maths is provided. A student can find 12th class maths chapter 2. anything and everything. Also, if both the students have any doubts in Mathematics, they can visit the website and ask their questions and download the NCERT Book Solutions for Plus Two Chapter 2 PDF Version. If in doubt they will come back to you and clarify in a good way.

Inverse trigonometry is often overlooked, students think that they will never have to use it in real life. Wrong! It can be used for everything from combination locks to an early warning system for wheelchairs. So get cracking and see if you can solve all 35 of the inverse trigonometry problems presented in this PDF.

 Board SCERT, Kerala Text Book NCERT Based Class Plus Two Subject Math's Textbook Solution Chapter Chapter 2 Exercise Ex 2.2 Chapter Name Inverse Trigonometric Functions Category Plus Two Kerala

## Kerala Syllabus Plus Two Math's Textbook Solution Chapter  2 Inverse Trigonometric Functions Exercises 2.2

### Chapter  2  Inverse Trigonometric Functions Solution

Kerala plus two maths NCERT textbooks, we provide complete solutions for the exercise and answers provided at the end of each chapter. We also cover the entire syllabus given by the Board of secondary education, Kerala state.

### Chapter  2  Inverse Trigonometric Functions Exercise   2.2

$cos^{-1}\left&space;(&space;cos\frac{7\pi&space;}{6}&space;\right&space;)$  is equal to A) $\frac{7\pi&space;}{6}$ B) $\frac{5\pi&space;}{6}$ C) $\frac{\pi&space;}{3}$ D) $\frac{\pi&space;}{6}$

We know that cos−1 (cos x) = x if, which is the principal value branch of cos −1x.

Here,

Now, can be written as:

cos1(cos7π6) = cos1[cos(π+π6)]cos1(cos7π6) = cos1[ cosπ6]             [as, cos(π+θ) =  cos θ]cos1(cos7π6)  = cos1[ cos(π5π6)]cos1(cos7π6) = cos1[{ cos (5π6)}]   [as, cos(πθ) =  cos θ]cos-1cos7π6 = cos-1cosπ+π6cos-1cos7π6 = cos-1- cosπ6             as, cosπ+θ = - cos θcos-1cos7π6  = cos-1- cosπ-5π6cos-1cos7π6 = cos-1-- cos 5π6   as, cosπ-θ = - cos θ

Find the values of each of the expressions

1. $sin^{-1}\left&space;(&space;sin\frac{2\pi&space;}{3}&space;\right&space;)$

2. $tan^{-1}\left&space;(&space;tan\frac{3\pi&space;}{4}&space;\right&space;)$

3. $tan\left&space;(&space;sin&space;^{-1}&space;\frac{3}{5}+cot^{-1}\frac{3}{2}\right&space;)$

1.

Ans.

$sin^{-1}sin\left&space;(&space;\pi&space;-\frac{\pi&space;}{3}&space;\right&space;)$

=$sin^{-1&space;}sin\left&space;(&space;\frac{\pi&space;}{3}&space;\right&space;)$

$=\frac{\pi&space;}{3}$

2.

Now, can be written as:

3.

Ans. $tan&space;\left&space;(&space;tan^{-1}\frac{3}{4}+&space;tan^{-1}&space;\frac{2}{3}\right&space;)$

$tan.&space;tan^{-1}\frac{\left&space;(&space;\frac{3}{4}+\frac{2}{3}&space;\right&space;)}{1-\frac{3}{4}\times&space;\frac{2}{3}}$

$=tan&space;.tan^{-1}&space;\frac{(9+8)}{6}$

$=\frac{17}{6}$

$tan^{-1}\sqrt{3}-cot^{-1}\left&space;(&space;-\sqrt{3}&space;\right&space;)$ is equal to A) $\pi$ B) $\frac{-\pi&space;}{2}$C) 0 D) $2\sqrt{3}$

Let. Then,

We know that the range of the principal value branch of

Let.

The range of the principal value branch of

$sin\left&space;(&space;\frac{\pi&space;}{3}-sin^{-1}\left&space;(&space;-\frac{1}{2}&space;\right&space;)&space;\right&space;)$ A) $\frac{1}{2}$ B) $\frac{1}{3}$ C) $\frac{1}{4}$ D) $1$

Since sin inverse of 0.5 is -30 degree. Since sin(60-(-30)) = sin 90 =1

Prove the following:

1. $3sin^{-1}x=sin^{-1&space;}(3x-4x^{3}),&space;x&space;\epsilon&space;\left&space;\lfloor&space;-\frac{1}{2}&space;,&space;\frac{1}{2}\right&space;\rfloor$

2. $3cos^{-1}x=cos^{-1}\left&space;(&space;4x^{3}&space;-3x\right&space;),&space;x\epsilon\left&space;[&space;\frac{1}{2},1&space;\right&space;]$

3. $tan^{-1}\left&space;[&space;\frac{2}{11}&space;\right&space;]+tan^{-1}\frac{7}{24}=tan^{-1}=\frac{1}{2}$

4. $2tan^{-1}\frac{1}{2}+tan^{-1}\frac{1}{7}=tan^{-1}\frac{31}{17}$

(1)To prove:

Let x = sinθ. Then,

We have,

R.H.S. =

= 3θ

= L.H.S.

(2)To prove:

Let x = cosθ. Then, cos−1x =θ.

We have,

(3)To prove:

(4)To prove:

Write the following functions in the simplest form:

1. $tan^{-1}=\frac{\sqrt{1+x^{2}}-1}{x},x\neq&space;0$ 2. $tan^{-1}\frac{1}{\sqrt{x^{2}-1}}\left&space;|&space;x&space;\right&space;|>&space;1$

3. $tan^{-1}&space;\left&space;(&space;\sqrt{\frac{1-cosx}{1+cosx}}&space;\right&space;),0<&space;x<&space;\pi$ 4. $tan^{-1}\left&space;(&space;\frac{cosx-sinx}{cosx+sinx}&space;\right&space;),-\frac{\pi&space;}{4}<&space;x<&space;\frac{3\pi&space;}{4}$

5. $tan^{-1}\frac{x}{\sqrt{a^{2}-x^{2}}},\left&space;|&space;x&space;\right&space;|<&space;a$  6. $tan^{-1}\left&space;(&space;\frac{3a^{2}x-x^{3}}{a^{3}-3ax^{2}}&space;\right&space;),&space;a>&space;0,&space;\frac{-a}{\sqrt{3}}<&space;x<&space;\frac{a}{\sqrt{3}}$

1.

2.

Put x = cosec θ ⇒ θ = cosec−1x

3.

4.

$tan^{-1}(\frac{cosx-sinx}{cosx+sinx})$

$=tan^{-1}(\frac{1-\frac{sinx}{cosx}}{1+\frac{sinx}{cosx}})$

$=tan^{-1}(\frac{1-tanx}{1+tanx})$

$=tan^{-1}(1)-tan^{-1}(tanx)(tan^{-1}\frac{x-y}{1+xy}=tan^{-1}x-tan^{-1}y)$

$=\frac{x}{4}-x$

5.

6.

Find the values of each of the following:

1. $tan^{-1}\left&space;[&space;2cos\left&space;(&space;2sin^{-1}&space;\frac{1}{2}\right&space;)&space;\right&space;]$                     2. $cot\left&space;(&space;tan^{-1}a+cot^{-1}&space;a\right&space;)$

3. $tan\frac{1}{2}\left&space;[&space;sin^{-1}\frac{2x}{1+x^{2}}&space;+cos^{-1}\frac{1-y^{2}}{1+y^{2}}\right&space;]$  , | x | < 1, y > 0 and xy < 1

4. If $sin\left&space;(&space;sin^{-1}&space;\frac{1}{5}+cos^{-1}x\right&space;)=1$ , then find the value of x

5. If $tan^{-1}\frac{x-1}{x-2}+tan&space;^{-1}\frac{x+1}{x+2}=\frac{\pi&space;}{4}$ , then find the value of x

(1) Let. Then,

(2)
(3)

Let x = tan θ. Then, θ = tan−1x.

Let y = tan Φ. Then, Φ = tan−1y.

(4)

On squaring both sides, we get:

Hence, the value of x is

(5)

Hence, the value of x is

#### Chapter 2: Inverse Trigonometric Functions EX 2.2 Solution- Preview

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