Inverse Trigonometric Functions NCERT Solutions

NCERT Miscelaaneous Exercise Q:9-17

Prove That:

Question:9 .

`tan^-1sqrtx=1/2cos^-1((1-x)/(1+x))` `x\ in[0,1]`

solution:

R.H.S.`= 1/2 cos^-1[(1-x)/(1+x)]`

let`x=tan^2theta,`then`sqrtx = tan theta`

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

`=1/2cos^-1(cos2theta)`

`=1/2 xx2theta =theta`

`=tan^-1sqrtx=L.H.S.`

Question:10 .

`cot^-1[(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx))]` `=x/2,x in(0,\  pi/4)`

solution:

L.H.S.`=cot^-1[(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx))]`

`=cot^-1[(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx))]`

`xx cot^-1[(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx))]`

`=cot^-1[((sqrt(1+sinx)+sqrt(1-sinx))^2)/((sqrt(1+sinx))^2-(sqrt(1-sinx))^2)]`

`=Cot^-1[(1+sinx+1-sinx+2sqrt(1-sin^2x))/(1=sinx-1sinx)]`

`=cot^-1[(2+2cosx)/(2sinx)]`

`=cot^-1[(1+cosx)/sinx]`

`=cot^-1[(2cos^2\ x/2)/(2sin\ x/2 cos\ x/2)]`

`=cot^-1[cot\ x/2]=x/2=`R.H.S.

Question:11 .

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

`pi/4-1/2 cos^-1x,``- 1/sqrt2 <= x <= 1`

solution:

L.H.S.`tan^-1[(sqrt(1+x)-sqrt(1-x))/(sqrt(1+x) +sqrt(1-x))]`

let `\ x=cos2theta,` then

LHS `=tan^-1[(sqrt(1+cos2theta)-sqrt(1-cos2theta))/(sqrt(1+cos2theta) +sqrt(1-cos2theta))]`

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

`=tan^-1[(costheta-sintheta)/(costheta+sintheta)]`

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

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

`=pi/4 -theta= pi/4 - 1/2 cos^-1 x` = R.H.S.

Question: 12 .

`(9pi)/8 -9/4 sin^-1\ 1/3 =9/4sin^-1\ (2sqrt2)/3`

solution:

`=>(9pi)/8 = 9/4 sin^-1\ 1/3 +9/4sin^-1\ (2sqrt2)/3`

R.H.S.`=9/4 sin^-1\ 1/3 +9/4sin^-1\ (2sqrt2)/3`

`=9/4[sin^-1\ 1/3+sin^-1\ (2sqrt2)/3]`

`=9/4[sin^-1(1/3sqrt(1-8/9)+(2sqrt2)/3sqrt(1-1/9))]`

[∵`sin^-1x+sin^-1y` `=sin^-1(xsqrt1-y^2+ysqrt(1-x^2)]`

`=9/4[sin^-1(1/3xx1/3+(2sqrt2)/3xx(2sqrt2)/3)]`

`=9/4sin^-1[1/9+8/9]=9/4sin^-1(1)`

`=9/4xxpi/2=(9pi)/8 =`R.H.S.

Question:13 .

`2tan^-1(cosx)=tan^-1(2cosec\ x)`

solution:

Here `\ 2tan^-1(cosx)=tan^-1(2cosec\ x)`

`=>tan^-1((2cosx)/(1-cos^2x))` `=tan^-1(2/(sinx))`

[∵`2tan^-1x=tan^-1\ (2x)/(1-x^2)]`

`=>(2cosx)/(sin^2x)=2/sinx`

`=>cotx=1=>x=cot^-1(1)`

`=>x=pi/4`

Question:14 .

`tan^-1\ (1-x)/(1+x)` `=1/2tan^-1x(x > 0)`

solution:

Here `tan^-1\ (1-x)/(1+x)` `=1/2tan^-1x`

`=>2tan^-1\ (1-x)/(1+x)= tan^-1x`

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

`=tan^-1[((2(1-x))/(1+x))/((4x)/(1+x)^2)]`

`=>tan^-1[(2(1-x))/(1+x)^2xx(1+x)^2/(4x)]=tan^-1x`

`=>(1-x^2)/(2x)= x =>1-x^2`

`=2x^2 => 3x^2=1 => x =+-1/sqrt3`

But `x> 0` So `\ x =1/sqrt3`

Question: 15 .

`sin(tan^-1x),|x| <1` is equal to

(A)`x/sqrt(1-x^2)`

(B)`1/sqrt(1-x^2)`

(C)`1/sqrt(1+x^2)`

(D)`x/sqrt(1+x^2)`

solution:

`sin(tan^-1x)` `=sin[sin^-1\ x/sqrt(1+x^2)]`

`= x/sqrt(1+x^2)`

[∵`tan^-1x = sin^-1\ x/sqrt(1+x^2) ]`

Thus answer is(D)

Question:16 .

`sin^-1(1-x)-2 sin^-1 x =pi/2`

then x is equal to

(A)`0,1/2`

(B)`1,1/2`

(C) 0

(D)`1/2`

solution:

Here `sin^-1(1-x)-2sin^-1 x=pi/2`

Let `x= sintheta`, then

`sin^-1(1-sintheta)- 2theta =pi/2`

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

`=>-2theta= cos^-1(-1 sintheta)`

`=>cos(-2theta)=1-sin theta`

`=>cos2theta = 1-sin theta`

`=>1-2sin^2theta = 1 - sintheta`

`=>2sin^2theta = sin theta`

`=>2sin^2 theta-sin theta=0`

`=>sin theta(2sintheta-1)=0`

Either `sintheta=0`or `2sintheta -1 = 0`

`:. x=0 or 2x-1 =0`

`=>x = 1/2`

Question:17 .

`tan^-1(x/y)-tan^-1((x-y)/(x+y))`

is equal to

(A)`pi/2`

(B)`pi/3`

(C)`pi/4`

(D)`-(3pi)/4`

solution:

Here `tan^-1(x/y)-tan^-1((x-y)/(x+y))`

`=tan^-1[(x/y-(X-Y)/(x+y))/(1+x/yxx(x-y)/(x+y))]`

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

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

`=tan^-1[(x^2+xy-xy+y^2)/(xy+y^2+x^2-xy)]`

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

Thus answer is (C)

12-math-home

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