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Math Twelve

Inverse Trigonometric Functions NCERT Solutions

NCERT Exercise 2.1

Find the principle values of the following

Question: 1. sin-1(-12)

Solution:

let sin-1(-12)=θ

sinθ=-12

Now we know that the range of principal value branch of sin-1 is [-π2,π2]

=-sinpi/6=sin(-pi/6)

=>theta=pi/6in[-pi/2,pi/2]

Thus, principal value of sin^-1(-1/2) is -pi/6

Question: 2. cos^-1(sqrt3/2)

Let cos^-1(sqrt3/2)=theta

=> sin theta = sqrt3/2

Now we know that the range of principal value branch of cos^-1 is [0,pi]

:. cos theta = sqrt3/2=cos pi/6

=> theta=pi/6 in [0, pi]

Question: 3. cosec^-1(2)

Let cosec^-1(2)=theta

Now we know that the range of principal value branch of cosec^-1 is [-pi/2, pi/2]-[0]

:. cosec theta = 2 = cosec pi/6

=> theta = pi/6 in [-pi/2, pi/2]-[0]

Thus principal value of cosec^-1(2) is pi/6

Question: 4. tan^-1(-sqrt3)

Let tan^-1(-sqrt3)=theta

=> tan theta =-sqrt3

We know that the range of principal value branch of tan^-1 is (-pi/2, pi/2)

:. tan theta =-sqrt3

=-tan pi/3 = tan (-pi/3)

=> theta = -pi/3 in (-pi/2, pi/2)

Thus, principal value of tan^-1(-sqrt3) is -pi/3

Question: 5. cos^-1(-1/2)

Solution: Let cos^-1 (-1/2) = theta

=> cos theta = -1/2

We know that the range of principalvalue of cos^-1 is [0, pi]

:. cos theta = -1/2=-cos pi/3

=cos(pi-pi/3) = cos 2pi/3

=> theta = 2pi/3 in [o, pi]

Thus principal value of cos^-1(-1/2) is 2pi/3

Question: 6.tan^-1(-1)

Solution:

let tan^-1(-1)=theta

=>tantheta=-1

we know that the range of principal value of tan^-1 is(-pi/2,pi/2)

:.tantheta=-1=-tan pi/4 =tan(-pi/4)

=>theta=- pi/4 in (- pi/2,pi/2)

Thus principal value of tan^-1 (-1) is - pi/4

Question: 7.sec^1(2/sqrt3)

Solution:

Let sec^-1(2/sqrt3)=theta

We know that the range of principal value branch of sec^-1 is [0,pi]-{pi/2}

:. sectheta=2/sqrt3 =sec(pi/6)

=>theta=pi/6in[0,pi]-{pi/2}

Thus principla value of sec^-1(2/sqrt3) is pi/6

Question:8 .cot^-1(sqrt3)

Solution:

Let cot^-1(sqrt3)=theta

=> cottheta =sqrt3

we know that the range of principal value branch of cot^-1 is (0.pi)

:. cottheta=sqrt3=cot(pi/6)

=> theta = pi/6in(0,pi)

Thus principal value of cot^1(sqrt3) is pi/6

Question:9 .cos^-1(- 1/sqrt2)

Solution:

Let cos^-1(- 1/sqrt2)=theta

=>costheta=- 1/sqrt2

we know that the range of principal value branch of cos^-1 is [0,pi]

:.costheta=-1/sqrt2 =-cos pi/4

=cos (pi -pi/4)=cos(3pi)/4

=>theta =(3pi)/4in[0,pi]

Thus principal value of cos^-1(- 1/sqrt2) is (3pi)/4

Question:10 .cosec^-1(-sqrt2)

Solution:

Let cosec^-1(-sqrt2)=theta

=>cosec theta= -sqrt2

we know that the range of principal value branch of cosec^-1 is [- pi/2,pi/2]-{0}

=> theta = -pi/4in[- pi/2,pi/2]-{0}

:. cosec theta = -sqrt2 =-cosec(pi/4)=cosec(- pi/4)

=>theta=- pi/4in[- pi/2,pi/2]-{0}

Thus principal value of cosec^-1(-sqrt2) is -pi/4

Find the value of the following:

Question:11 .

tan^-1(1)+cos^-1(- 1/2)+ sin^-1(- 1/2)

Solution:

Let tan^-1(1)= theta_1

=>tan theta_1 = 1 tan (pi/4)

theta_1=pi/4in(- pi/2,pi/2)

Let cos^-1(-1/2)=theta_2

=>Costheta_2=-1/2 = -cos(pi/3) =cos(pi-pi/3)=cos (2pi)/3

=>theta_2=(2pi)/3 in [0,pi]

Let sin^-1(- 1/2)=theta_3

=> sintheta_3=- 1/2 =-sin (pi/6)=sin(- pi/6)

=>theta_3= - pi/6in [- pi/2,pi/2]

:. tan^-1 (1)+ cos^-1(- 1/2)+ sin^-1(- 1/2) =theta_1+theta_2+theta_3=pi/4+(2pi)/3-pi/6

=(3pi+8pi-2pi)/12 =(9pi)/12 =(3pi)/4

Question:12 . cos^-1(1/2)+2sin^-1(1/2)

Solution:

Let cos^-1(1/2)=theta_1

=> cos theta_1=1/2=cospi/3

=>theta_1=pi/3in[o,pi]

Let sin^-1 (1/2)=theta_2

=>sintheta_2=1/2=sin pi/6

theta_2 = pi/6in[- pi/2,pi/2]

:.cos^-1(1/2)+2sin^-1(1/2) =theta_1+theta_2=pi/3+(2xpi)/6=(2pi)/3

Question:13 .if sin^-1x=y then

(A) 0 <= y <= pi

(B) -pi/2 <= y <= pi/2

(C) 0 < y < pi

(D) -pi/2 < y < pi/2

Solution:

sin^-1 x =y => sin y = x

We know that the range of principal value branch of sin^-1 is [-pi/2, pi/2]

So y in [-pi/2, pi/2]

Thus answer is (B) -pi/2 <= y <= pi/2

Question: 14. tan^-1 sqrt3 - sec^-1(-2) is equal to

(A) pi

(B) -pi/3

(C) pi/3

(D) (2pi)/3

Let tan^-1 sqrt3 =theta_1

=> tan^-1 sqrt3 = tan (pi/3)

=> theta_1 = pi/3 in (-pi/2, pi/2)

Let sec^-1(-2) = theta_2

=> sec theta_2 = -2 = -sec (pi/3)

=sec (pi-pi/3)= sec 2pi/3

=> theta_2 = (2pi)/3 in [o, pi]-(pi/2)

:. tan^-1 sqrt3 - sec^-1(-2) = theta_1-theta_2

= pi/3 - (2pi)/3 = -pi/3

Thus answer is (B) -pi/3

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