Math Twelve

Inverse Trigonometric Functions NCERT Solutions

NCERT Exercise 2.2

Prove the following

Question:1 .

`3sin^-1x=sin^-1(3x-4x^3), ` `x` ` in[- 1/2,1/2]`

Solution:

`3sin^-1x=sin^-1(3x-4x^3), ` `x` ` in[- 1/2,1/2]`

Let `x=sin theta` then

L.H.S =`3 sin^-1 x`

=`3sin^-1(sin theta)=3 theta`

R.H.S =`sin^-1 (3x - 4x^3)`

` =sin^-1[3 sin theta - 4sin^3 theta]`

`=sin^-1[sin 3theta]` `[:.sin 3theta =3 sintheta - 4 sin^3 theta]`

`=3theta`

`:.L.H.S.=R.H.S.`

Question:2 .

`3 cos^-1 x` ` =cos^-1(4x^3-3x)` `x` `in [-1/2,1]`

Solution:

Here `3 cos^-1 x` ` =cos^-1(4x^3-3x)` `x` `in [-1/2,1]`

Let `x=costheta,then`

L.H.S. `=3cos^-1 x`

`=3 cos^-1(cos theta) =3 theta`

R.H.S. `=cos^-1(4x^3-3x)`

`=cos^-1[4cos^3 theta-3cos theta]`

`=cos^-1[cos 3 theta]` `[:. cos 3theta = costheta - 4 cos^3 theta]`

`=3theta`

`=L.H.S. = R.H.S.`

Question:3 .

`tan ^-1 (2/(11))+ tan^-1 (7/(24))` `=tan^-1 (1/2)`

Solution:

Here `tan ^-1 (2/(11))+ tan^-1 (7/(24))` `=tan^-1 (1/2)`

L.H.S.`= tan^-1\ 2/11=tan^-1\ 7/24`

`tan^-1 ((2/11+7/24))/(1-2/11xx 7/24)`

[∵ `tan^-1\ x+tan^-1\ y=``tan^1\ (x+y)/(1-xy)`]

`=tan^-1(((48+77)/(264))/(1- 14/264))`

`=tan^-1((125/264)/((264-14)/(264)))`

`=tan^-1((125/264)/(250/264))`

`tan^-1(125/264xx264/250)`

`tan^-1\ 1/2 = R.H.S.`

Question:4 .

`2tan^-1\ 1/2 + tan^-1\ 1/7 = tan^-1\ 31/17`

Solution:

Here `2tan^-1\ 1/2 + tan^-1\ 1/7 = tan^-1\ 31/17`

L.H.S.`= tan^-1\ 1/2 + tan^-1\  1/7`

`=tan^-1\ (2xx1/2)/(1-(1/2)^2)+tan^-1\ 1/7`

[∵ `2tan^-1\ x = tan^-1\ (2x)/(1-x^2)-1 < x < 1`]

`=tan^-1\ 1/(1-(1/4))+ tan^-1\ 1/7`

`=tan^-1\  4/3 + tan^-1\ 1/7`

`= tan^-1\ ((4/3+1/7))/(1- 4/3xx1/7)`

[∵ `tan^-1\ x +tan^-1\ y= tan^-1\ (x+y)/(1-xy)`]

` tan^-1\ ((28+3)/21)/(1- (4/21))`

`tan^-1\  (31/21)/(17/21)`

`tan^-1\ 31/21 xx 21/17`

`=tan^-1\ 31/17 =R.H.S.`

Write the follwing functions in the simplest from.

Question:5 . `tan^-1 (sqrt1+x^2-1)/x x !=0`

Solution:

Here `tan^-1* (sqrt1+x^2-1)/x x !=0`

let`x=tan theta`, then

`tan^-1*(sqrt1+x^2-1)/x`

` =tan^-1[(sqrt1+tan^2theta-1)/tan theta]`

`=tan^-1 [(sec theta-1)/tan theta]`

`=tan^-1[(1/cos theta -1)/(sin theta/cos theta)]`

`=tan^-1[(1-cos theta)/sin theta]`

`=tan^-1[(2 sin^2* theta/2)/(2sin* theta/2 cos* theta/2)]`

[∵ `cos 2 theta= 1-2 sin^2theta,`]

`sin 2 theta =2 sin theta cos theta]`

`=tan^-1[(sin* theta/2)/(cos* theta/2) ]`

`tan^-1(tan*theta/2)`

`= theta/2= 1/2*tan^1 x`

`thus tan^-1*(sqrt1+x^2-1)/x = 1/2 *tan^-1x `

Question:6 .`tan^-1* 1/(sqrtx^2-1),|x| >1`

Solution:

`tan^-1* 1/(sqrtx^2-1),|x| >1`

`letx=sec theta,` then

`tan^-1* 1/(sqrtx^2-1)`

`=tan^-1* 1/(sqrtsec^2 theta-1)`

`tan^-1*(1/tantheta)=tan^-1 (cot theta)`

`=tan^-1[tan(pi/2-theta)]=pi/2-theta`

[∵ `theta =sec^-1 x`]

`=pi/2-sec^-1x`

Thus `tan^-1*1/sqrt(x^2-1)=pi/2-sec^-1x`

Question: 7 .

`tan^-1(sqrt((1-cosx)/(1+cosx))) x < pi`

Solution:

Here `tan^-1(sqrt((1-cosx)/(1+cosx))) x < pi`

`tan^-1(sqrt((1-cosx)/(1+cosx)))`

`=tan^-1(sqrt((2sin^2*x/2)/(2cos^2*x/2)))`

[∵ `cos 2 theta= 2 cos^2theta-1` `=1-2sin^2 theta`]

`=tan^-1(sqrt(tan^2*x/2)`

`=tan^-1(tan*x/2)=x/2`

Thus `tan^-1(sqrt((1-cosx)/(1+cosx))) = x/2`

Question: 8.

`tan^-1((cosx - sinx)/(cosx+sinx)),0 < x < pi`

Solution:

Here, `tan^-1((cosx - sinx)/(cosx+sinx)),0 < x < pi`

Dividing numerator and denominator by cos x

`:.tan^-1((cosx - sinx)/(cosx+sinx)),0 < x < pi`

`tan^-1*(((cosx - sinx)/(cosx+cosx))/((cosx - sinx)/(cosx+cosx)))`

`=tan^-1* [(1-tan x)/(1+tan x)]`

`= tan^-1* [tan(pi/4 -x)]= pi/4-x `

`Thus, tan^-1*((cosx-sinx)/(cosx+sinx))= pi/4 - x `

Question:9 .

`tan^-1*x/(sqrta^2-x^2) |x| < a`

Solution:

Here`tan^-1*x/(sqrta^2-x^2) |x| < a`

Let `x = a sin theta`, then

`tan^-1*x/(sqrta^2-x^2)`

`tan^-1*[(a sin theta)/(sqrta^2-a^2sin^2 theta)] `

` tan^-1*[(a sin theta)/(a cos theta)]`

`-tan^-1(tan theta) = theta`

`=sin^-1* (x/a)` [∵ `theta = sin^-1*x/2`]

Thus, `tan^-1* x/ (sqrt x^2-a^2)= sin^-1 (x/a)`

Question:10 .

`tan^-1*[(3a^2x-x^3)/(a^3-3ax^2)],` `a > 0; -a/sqrt3 <= x <= a/sqrt3`

Solution:

Here, `tan^-1*[(3a^2x-x^3)/(a^3-3ax^2)],` `a > 0; -a/sqrt3 <= x <= a/sqrt3`

Let `x = atan theta`, then

`tan^-1*[(3a^2x^2-x^3)/(a^3-3ax^2)]`

` tan^-1*[(3a^2xxatan theta-a^3tan^3 theta)/(a^3-3axxa^2 tna^2 theta)]`

`tan^-1*[(3tan theta - tan^3 theta)/(1- 3tan^2theta)]`

`=tan^-1(tan3theta)` `[:. tan3theta = (3tan theta -tan^3theta)/(1-3tan^2 theta) ]`

[∵ `theta = tan^-1 x/a`]

`=3theta = 3 tan^-1 (x/a)`

Thus `tan^-1*[(3a^2x-x^3)/(a^3-3ax^2)]`

`=3 tan^-1*(x/a)`

Find the values of each of the following

Question: 11 .

`tan^-1[2cos(2sin^-1*1/2)]`

Solution:

Let `sin^-1(1/2)=theta` ` => sin theta = 1/2`

:.`tan^-1[2cos(2sin^-1*1/2)]`

`= tan^-1[2 cos2 theta]`

`= tan^-1[2(1-2sin^2 theta)]`

[∵ `cos 2theta = 1 - 2 sin^2 theta`]

`=tan^-1[2(1-2xx(1/2)^2)]`

`tan^-1[2(1-1/2)]`

`tan^-1(1)=tan^-1(tan*pi/4)=pi/4`

Thus `tan^-1[2cos (2sin^-1*1/2)]=pi/4`

12-math-home


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