Inverse Trigonometric Functions NCERT Solutions
NCERT Exercise 2.2 Q: 12 to 21
Question:12 .`cot(tan^-1a+cot^-1a)`
Solution: `cot (tan^-1a+cot^-1a)=cot (pi/2)`
`[:.tan^-1x+cot^-1x=pi/2,x in R]`
`= 0 `
Question:13 .
`tan*1/2[sin^-1* (2x)/(1+x^2)+cos^-1*(1-y^2)/(1+y^2)],`
`|x| < 1, y > 0 and xy < 1.`
Solution:
`tan*1/2[sin^-1* (2x)/(1+x^2)+cos^-1*(1-y^2)/(1+y^2)],`
`=tan*1/2[2 tan^-1 +2 tan^-1y]`
[∵ `2 tan^-1x=sin^-1*(2x)/(1+x^2)` and
`2tan^-1y=cos^-1*(1-y^2)/(1+y^2)]`
`tan*1/2xx2[tan^-1x=tan^-1y]`
`=tan*[tan^-1*(x+y)/(1-xy)]`
[∵ `tan^-1x+tan^-1y` `=tan^-1* (x+y)/(1-xy)`]
`=(x+y)/(1-xy)`
Question:14 .If `sin(sin^-1*1/5 +cos^-1 x)=1` then find the value of `x`
Solution:
`sin(sin^-1*1/5 +cos^-1 x)=1`
`=>[sin^-1*1/5+sin^-1*sqrt(1-x^2)]=1`
[∵ `cos^-1x` `=sin^-1sqrt(1-x^2)`]
`=>sin[sin^-1(1/5sqrt(1-1+x^2)` `+sqrt(1-x^2)sqrt (1-(1/25)))]=1`
[∵`sin^-1x+sin^-1y` `=sin^-1[xsqrt(1-y^2) + ysqrt(1-x^2)]`]
`=>sin[sin^-1(1/5*x` `+sqrt(24/25) sqrt(1-x^2))]=1`
`=>1/5*x+sqrt(24/25)sqrt(1-x^2)=1`
`=>sqrt(24/25)sqrt(1-x^2)=1-x/5`
Squaring both sides, we have
`24/25(1-x^2)=(1-x/5)^2`
`=>24-24x^2` `=25(1+x^2/25 -(2x)/5)`
`=>24-24x^2` `=25+x^2-10x`
`=>25x^2-10x+1=0` `=>25x^2-5x-5x+1=0`
`=>(5x-1)^2 =0` ` =>5x-1 = 0 => x = 1/5`
Thus` x=1/5 ` is a root of given equation.
Question:15 .
If `tan^1 *(x-1)/(x-2) + tan^-1* (x+1)/(x+2)` `=pi/4` then find the value of x.
Solution:
Here `tan^-1*(x-1)/(x-2)` `+tan^-1*(x+1)/(x+2)=pi/4`
`=>tan^-1*(x-1)/(x-2)` `+tan^-1*(x+1)/(x+2)=tan^-1(1)`
`tan^-1*(x-1)/(x-2)=tan^-1(1)` `-tan^-1*(x+1)/(x+2)`
`tan^-1*(x-1)/(x-2)` `=tan^-1*[1-((x+1)/(x+2))/(1+(x+1)/(x+2))]`
[∵`tan^-1x -tan^-1y` `=tan^-1*(x-y)/(1+xy)`]
`=>tan^-1*(x-1)/(x-2)` `=tan^-1[(x+2-x-1)/(x+2+x+1)]`
`=>tan^-1*(x-1)/(x-2)` `tan^-1[1/(2x+3)]`
`(x-1)/(x-2)= 1/(2x+3)`
`=>(x-1)(2x+3)=x-2`
`=>2x^2+x-3=x-2` `=>2x^2-1=0` `=>x^2=1/2 =>x=+- 1/sqrt2`
Thus `x=+- 1/sqrt2` is a root of given equation.
find the values of each of the expressions inExercises 16 to 18
Question:16 . `sin^-1(sin*(2pi)/3)`
Solution:Here `sin^-1(sin*(2pi)/3)`
`sin^-1*[sin*(pi-2pi/3)]`
Now `pi/3 in [-pi/3,pi/2]`
`:. sin^-1(sin*(2pi)/3)`
`=sin^-1*(sin*pi/3)=pi/3`
Thus, `sin^-1(sin*(2pi)/3)=pi/3`
Question:17 .`tan^-1(tan*(3pi)/4)`
Solution:Here `tan^-1(tan*(3pi)/4)`
`=>tan^-1[tan(pi*-pi/4)]`
Now `pi/4 in (-pi/2,pi/2)`
`:.Tan^-1(tan*3pi/4)` `=tan^-1[tan(-*pi/4)]=-*pi/4`
Thus `tan^-1(tan*3pi/4)=pi/4`
Question:18 .
`tan[sin^-1*3/5+cot^-1*3/2]`
Solution:Here `tan[sin^-1*3/5+cot^-1*3/2]`
`=>tan[tan^-1* (3/5)/sqrt(1-(3/5)^2) +tan^-1*2/3]`
[∵ `sin^-1x=tan^-1*x/sqrt(1-x^2)`]
`and cot^-1x = tan^-1*1/x]`
`tan[tan^-1*3/4 +tan^-1*2/3]`
`tan[tan^-1*((3/4+2/3))/(1-3/4xx2/3)]`
`=tan[tan^-1*(17/12)/(1/2)]`
`=tan[tan^-1*17/6]=17/6`
Thus `tan[sin^-1*3/5+cot^-1*3/2]` `=17/6`
Question:19 .
`cos^-1(cos*(7pi)/6)` is equal to
(A)`(7pi)/6`
(B)`(5pi)/6`
(C)`pi/3`
(D)`pi/6`
Solution;
`cos^-1(cos*(7pi)/6)`
`=cos^-1[cos(2pi-(5pi)/6)]`
Now `5pi/6in[0,pi]`
`cos^-1(cos*(7pi)/6)` `=cos^-1(cos*(5pi)/6)=(5pi)/6`
Thus answer is (B)
Question:20 .
`sin[pi/3 -sin^-1(-*1/2)]` `is equal to
(A)`1/2`
(B)`1/2`
(C)`1/2`
(D)`1/2`
Solution:
`sin[pi/3 -sin^-1(-*1/2)]`
`=sin[pi/3+sin^-1*1/2]`
`=sin[pi/3+pi/6]`
`sin*pi/2=1`
Thus answer is (D)
Question:21 .
`tan^-1*sqrt3-cot^-1(-sqrt3)` is equal to
(A)`pi`
(B)`-*pi/2`
(C)0
(A)`2sqrt3`
Solution:
`tan^-1*sqrt3-cot^-1(sqrt3)`
`=tan^-1*sqrt3-(pi-cot^-1sqrt3)`
(∵`cot^-1(-x)=pi-cot^-1x`)
`=tan^-1*sqrt3- pi+cot^-1sqrt3`
`=tan^-1*sqrt3-pi+tan^-1*1/sqrt3`
(∵ `cot^-1x=tan^-1* 1/x`)
`(tan^-1sqrt3+tan^-1*1/sqrt3)-pi`
`tan^-1*(sqrt3+1/sqrt3)/(1-sqrt3xx1/sqrt3)-pi` `tan^-1(oo)-pi`
(∵`tan*pi/2=oo`)
`=pi/2-pi= -pi/2`
thus answer is(B).
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