# Plus Two Math's Solution Ex 2.1 Chapter2 Inverse Trigonometric Functions

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Here is the solution for Exercise 2.1 Chapter 2 Inverse Trigonometric Functions of NCERT plus two maths. Here we have given a detailed explanation of each and every exercise so that students can understand the concepts easily without any difficulty. The solution to each and every question is provided here so you can solve them by yourself if you don’t get the answer here.

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 Board SCERT, Kerala Text Book NCERT Based Class Plus Two Subject Math's Textbook Solution Chapter Chapter 2 Exercise Ex 2.1 Chapter Name Inverse Trigonometric Functions Category Plus Two Kerala

## Kerala Syllabus Plus Two Math's Textbook Solution Chapter  2 Inverse Trigonometric Functions Exercises 2.1

### 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   2.1

Question 1:

Find the principal values of the following:

(i) $sin^{-1}&space;\left&space;(&space;\frac{-1}{2}&space;\right&space;)$ (ii) $cos^{-1}\left&space;(&space;\frac{\sqrt{3}}2{}&space;\right&space;)$ (iii) $cosec^{^{-1}}(2)$

(iv) $tan^{-1}(-\sqrt{3})$ (v) $cos^{-1}\left&space;(&space;-\frac{1}{2}&space;\right&space;)$ (vi) $tan^{-1}(-1)$

(vii) $sec^{-1}\left&space;(&space;\frac{2}{\sqrt{3}}&space;\right&space;)$ (viii) $cot^{-1}\left&space;(&space;\sqrt{3}&space;\right&space;)$ (ix) $cos^{-1}\left&space;(-&space;\frac{1}{\sqrt{2}}&space;\right&space;)$

(x) $cosec^{-1}\left&space;(&space;-\sqrt{2}&space;\right&space;)$

Solution 1:

(i) Let sin-1 (-1/2) = y . then sin y=-1/2 = -sin(Π/6) = sin (-Π/6)

Implies the range of the principal value branch of sin-1 is [-Π/2,Π/2]

and sin (-Π/6) = -1/2 implies the principal value of sin-1(-1/2)= -Π/6

(ii)

Let cos-1(√3/2) = y . then cosy=√3/2 = cos(Π/6)

Implies the range of the principal value branch of  cos-1 is [0,Π]

and cos(Π/6) = √3/2 implies the principal value of cos-1(√3/2)= Π/6

(iii)

Let cosec-1(2) = y . then cosec y=2 = cosec(Π/6)

Implies the range of the principal value branch of  cosec-1 is  [-Π/2,Π/2]-{0}

and cosec(Π/6) = 2 implies the principal value of cosec-1(2)= Π/6

(iv)

Let tan-1(-√3) = y . then tan y=-√3 = -tan(Π/3) =tan(-Π/3)

Implies the range of the principal value branch of  tan-1 is  (-Π/2,Π/2)

and tan(-Π/3) = -√3 implies the principal value of tan-1(-√3)= -Π/3

(v)

Let cos-1(-1/2) = y . then cosy=-1/2 = -cos(Π/3) =cos(Π-Π/3) =cos(2Π/3)

Implies the range of the principal value branch of  cos-1 is [0,Π]

and cos(2Π/3) = -1/2 implies the principal value of cos-1(-1/2)= Π/6

(vi)

Let tan-1(-1) = y . then tan y=-1 = -tan(Π/4) =tan(-Π/4)

Implies the range of the principal value branch of  tan-1 is  (-Π/2,Π/2)

and tan(-Π/4) = -1 implies the principal value of tan-1(-1)= -Π/4

(vii)

Let sec-1(2/√3) = y . then sec y=2/√3 = sec(Π/6)

Implies the range of the principal value branch of  cosec-1 is  [-0,Π]-{Π/2}

and sec(Π/6)  = 2/√3 implies the principal value of sec-1(2/√3)= Π/6

(viii)

Let cot-1(√3) = y . then cot y=√3 = cot(Π/6)

Implies the range of the principal value branch of  cot-1 is  (0,Π)

and cot(Π/6) = √3 implies the principal value of cot-1(√3)= Π/6

(ix)

Let cos-1(-1/√2) = y . then cosy=-1/√2 = -cos(Π/4) =cos(Π-Π/4) =cos(3Π/4)

Implies the range of the principal value branch of  cos-1 is [0,Π]

and cos(3Π/4) = -1/√2 implies the principal value of cos-1(-1/√2)= 3Π/4

(x)

Let cosec-1(-√2) = y . then cosec y=-√2 = -cosec(Π/4) =cosec(-Π/4)

Implies the range of the principal value branch of  cosec-1 is  [-Π/2,Π/2]-{0}

and cosec(-Π/4) = -√2 implies the principal value of cosec-1(-√2)= -Π/4

Question 2:

Find the values of the following:

1. $tan^{-1}(1)+cos^{-1&space;}-\frac{1}{2}+sin^{-1}-\frac{1}{2}$

2. $cos^{-1}\frac{1}{2}+2sin^{-1}\frac{1}{2}$

Solution 2:

Question 3:

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

Solution 3:

Let tan$^{-1}\sqrt{3}=&space;x$. Then tan x $=\sqrt{3}=tan\frac{\pi&space;}{3}$

The range of the principle value branch of $tan^{-1}$ is $\left&space;[&space;\frac{-\pi&space;}{2}&space;,&space;\frac{\pi&space;}{2}\right&space;]$

$\therefore&space;tan^{-1}\sqrt{3}=&space;\frac{\pi&space;}{3}$

Let $sec^{-1}(-2)=y$ Then, sec y= $-2=-sec\left&space;(&space;\frac{\pi&space;}{3}&space;\right&space;)=&space;sec\left&space;(&space;\pi&space;-\frac{\pi&space;}{3}&space;\right&space;)=sec\frac{2\pi&space;}{3}$

We know that the range of the principal value branch of $sec^{-1}$ is $\left&space;[&space;0,&space;\pi&space;\right&space;]-&space;\left&space;\{&space;\frac{\pi&space;}{2}&space;\right&space;\}$

$\therefore&space;sec^{^{-1}}(-2)=\frac{2\pi&space;}{3}$

Hence $tan^{-1}\left&space;(&space;\sqrt{3}&space;\right&space;)-sec^{-1}(-2)$ = $\frac{\pi&space;}{3}-\frac{2\pi&space;}{3}=-\frac{\pi&space;}{3}$

Hence option B

Question 4:

If $sin^{-1}x=y$ then, A) $0\leq&space;y\leq&space;\pi$ B) $-\frac{\pi&space;}{2}\leq&space;y\leq&space;\frac{\pi&space;}{2}$

C) 0 < y < π D) $-\frac{\pi&space;}{2}<&space;y<&space;\frac{\pi&space;}{2}$

Solution 4:

It is given that sin−1y.

We know that the range of the principal value branch of sin−1 is

Therefore,.

#### Chapter 2: Inverse Trigonometric Functions EX 2.1 Solution- Preview

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