NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.1 - Inverse Trigonometric Functions

Question:1 Find the principal values of the following : sin−1⁡(−12)

Answer:

Let x=sin−1⁡(−12)

⟹sin⁡x=−12=−sin⁡(π6)=sin⁡(−π6)
We know, principle value range of sin−1 is [−π2,π2]

∴ The principal value of sin−1⁡(−12) is −π6,

Question:2 Find the principal values of the following: cos−1⁡(32)

Answer:

So, let us assume that cos−1⁡(32)=x then,

Taking inverse both sides we get;

cos x=(32) , or

cos(π6)=(32)

and as we know that the principal values of cos−1 is from [0, π ],

Hence cos x=(32)

when x = π6 .

Therefore, the principal value for cos−1⁡(32) is π6 .

Question:3 Find the principal values of the following: cosec−1(2)

Answer:

Let us assume that cosec−1(2)=x , then we have;

Cosec x=2 , or

Cosec(π6)=2 .

And we know the range of principal values is [−π2,π2]−{0}.

Therefore the principal value of cosec−1(2) is π6 .

Question:4 Find the principal values of the following: tan−1⁡(−3)

Answer:

Let us assume that tan−1⁡(−3)=x , then we have;

tan⁡x=(−3) or

−tan⁡(π3)=tan⁡(−π3).

and as we know that the principal value of tan−1 is (−π2,π2) .

Hence the only principal value of tan−1⁡(−3) when x=−π3 .

Question:5 Find the principal values of the following: cos−1⁡(−12)

Answer:

Let us assume that cos−1⁡(−12)=y then,

Easily we have; cos⁡y=(−12) or we can write it as:

−cos⁡(π3)=cos⁡(π−π3)=cos⁡(2π3).

as we know that the range of the principal values of cos−1 is [0,π] .

Hence 2π3 lies in the range it is a principal solution.

Question:6 Find the principal values of the following : tan−1⁡(−1)

Answer:

Given tan−1⁡(−1) so we can assume it to be equal to 'z';

tan−1⁡(−1)=z ,

tan⁡z=−1

or

−tan⁡(π4)=tan⁡(−π4)=−1

And as we know the range of principal values of tan−1 from (−π2,π2) .

As only one value z = −π4 lies hence we have only one principal value that is −π4 .

Question:7 Find the principal values of the following : sec−1⁡(23)

Answer:

Let us assume that sec−1⁡(23)=z then,

we can also write it as; sec⁡z=(23) .

Or sec⁡(π6)=(23) and the principal values lies between [0,π]−{π2} .

Hence we get only one principal value of sec−1⁡(23) i.e., π6 .

Question:8 Find the principal values of the following: cot−1⁡(3)

Answer:

Let us assume that cot−1⁡(3)=x , then we can write in other way,

cot⁡x=(3) or

cot⁡(π6)=(3) .

Hence when x=π6 we have cot⁡(π6)=(3) .

and the range of principal values of cot−1 lies in (0,π) .

Then the principal value of cot−1⁡(3) is π6

Question:9 Find the principal values of the following: cos−1⁡(−12)

Answer:

Let us assume cos−1⁡(−12)=x ;

Then we have cos⁡x=(−12)

or

−cos⁡(π4)=(−12) ,

cos⁡(π−π4)=cos⁡(3π4) .

And we know the range of principal values of cos−1 is [0,π] .

So, the only principal value which satisfies cos−1⁡(−12)=x is 3π4 .

Question:10 Find the principal values of the following: cosec−1(−2)

Answer:

Let us assume the value of cosec−1(−2)=y , then

we have cosec y=(−2) or

−cosec (π4)=(−2)=cosec (−π4) .

and the range of the principal values of cosec−1 lies between [−π2,π2]−{0} .

hence the principal value of cosec−1(−2) is −π4 .

Question:11 Find the values of the following: tan−1⁡(1)+cos−1⁡(−12)+sin−1⁡(−12)

Answer:

To find the values first we declare each term to some constant ;

tan−1(1)=x , So we have tan⁡x=1 ;

or tan⁡(π4)=1

Therefore, x=π4

cos−1(−12)=y

So, we have

cos⁡y=(−12)=−cos⁡(π3)=cos⁡(π−π3)=cos⁡(2π3) .

Therefore y=2π3 ,

sin−1⁡(−12)=z ,

So we have;

sin⁡z=−12 or

−sin⁡(π6)=sin⁡(−π6)=−12

Therefore z=−π6

Hence we can calculate the sum:

=π4+2π3−π6

=3π+8π−2π12=9π12=3π4 .

Question:12 Find the values of the following: cos−1⁡(12)+2sin−1⁡(12)

Answer:

Here we have cos−1⁡(12)+2sin−1⁡(12)

let us assume that the value of

cos−1⁡(12)=x,andsin−1⁡(12)=y ;

then we have to find out the value of x +2y.

Calculation of x :

⇒cos−1⁡(12)=x

⇒cos⁡x=12

⇒cos⁡π3=12 ,

Hence x=π3 .

Calculation of y :

⇒sin−1⁡(12)=y

⇒sin⁡y=12

⇒sin⁡π6=12 .

Hence y=π6 .

The required sum will be = π3+2(π6)=2π3 .

Question:13 If sin−1⁡x=y then

(A) 0≤y≤π

(B) −π2≤y≤π2

(C) 0<y<π

(D) −π2<y<π2

Answer:

Given if sin−1⁡x=y then,

As we know that the sin−1 can take values between [−π2,π2].

Therefore, −π2≤y≤π2 .

Hence answer choice (B) is correct.

Question:14 tan−1⁡(3)−sec−1⁡(−2) is equal to

(A) π

(B) −π3

(C) π3

(D) 2π3

Answer:

Let us assume the values of tan−1⁡(3) be 'x' and sec−1⁡(−2) be 'y'.

Then we have;

tan−1⁡(3)=x or

tan⁡x=3 or

tan⁡π3=3 or

x=π3 .

and sec−1⁡(−2)=y or

sec⁡y=−2

or −sec⁡(π3)=sec⁡(π−π3)=sec⁡2π3

y=2π3

also, the ranges of the principal values of tan−1 and sec−1 are (−π2,π2) . and

[0,π]−{π2} respectively.

∴ we have then;

tan−1⁡(3)−sec−1⁡(−2)

=π3−2π3=−π3