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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