Inverse Trigonometric Functions NCERT Solutions
NCERT Exercise 2.1
Find the principle values of the following
Question: 1. `sin^-1(-1/2)`
Solution:
let `sin^-1(-1/2)=theta `
`=> sin theta=-1/2`
Now we know that the range of principal value branch of `sin^-1` is `[-pi/2,pi/2]`
`:.sintheta=-1/2` `=-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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