Inverse Trigonometric Functions NCERT Solutions
NCERT Micellaneous Exercise
Find the values of the follwing
Question:1 .`cos^-1(cos*(13pi)/6)`
Solution:
Given, `cos^-1(cos*(13pi)/6)`
`=cos^-1[cos(2pi+ pi/6)]`
Now `pi/6in[0,pi]`
`:.cos^-1(cos*(13pi)/6)=`
`=cos^-1[cos*pi/6]=pi/6`
Question:2 .
`tan^-1(tan*(7pi)/6)`
Solution:`tan^-1(tan*(7pi)/6)`
`=tan^-1[tan(pi+pi/6)]`
Now `pi/6 in (- pi/2,pi/2)`
`:.tan^-1(tan*(7pi)/6)`
`=tan^-1[tan*pi/6]=pi/6`
Prove that
Question:3 .
` 2 sin^-1*3/5= tan^-1*24/7`
Solution:
L.H.S. `=2 sin^-1*3/5`
`=sin^-1[2xx3/5sqrt(1-(3/5))^2]`
[∵ `2sin^-1x=sin^-1(2xsqrt(1-x^2))]`
`sin^-1[2xx3/5xx4/5]`
`=sin^-1(24/25)`
`tan^-1[(24/25)/sqrt(1-(24/25)^2)]`
[∵ `sin^-1x=` `tan^-1*x/sqrt(1-x^2)]`
`tan^-1[(24/25)/sqrt(1-(576/625))]`
`=tan^-1[24/25xx25/7]`
`tan^-1[24/7]=R.H.S.`
Question:4 .
`sin^-1*8/17+sin^-1*3/5` `=tan^-1*77/36`
Solution:
L.H.S.=`sin^-1*8/17+sin^-1*3/5`
`sin^-1[8/17sqrt(1-(3/5)^2) ` `+3/5sqrt(1-(8/17)^2)]`
[∵ `sin^-1x+sin^-1y` `=sin^-1(xsqrt(1-y^2)+ysqrt(1-x^2))]`
`=sin^-1[8/17xx4/5+3/5xx15/17]`
`=sin^-1[32/85+45/85]` `=sin^-1[77/85]`
`tan^-1[(77/85)/sqrt(1-(77/85)^2)]`
[∵ `sin^-1x=` `tan^-1*x/sqrt(1-x^2)]`
`tan^-1[(77/85)/(36/85)]` `tan^-1*77/36=`R.H.S.
Question:5 .`cos^-1*4/5+cos^-1*12/13 =` ` cos^-1*33/65`
Solution:
L.H.S.`cos^-1*4/5+cos^-1*12/13 =`
`= cos^-1[4/5xx12/13-sqrt(1-(4/5)^2)` `sqrt(1-(12/13)^2)`
[∵ `cos^-1x+cos^-1y` `=cos^-1[xy-sqrt(1-x^2)sqrt(1-y^2]]]`
`=cos^-1[48/65-3/5xx5/13]`
`=cos^-1[48/65-3/13]` `=cos^-1*33/65=`R.H.S.
question:6 .
`cos^-1*12/13+Sin^-1*3/5` `=sin^-1*56/65`
Solution:
L.H.S.`cos^-1*12/13+Sin^-1*3/5` `=sin^-1sqrt(1-(12/13)^2)+sin^-1*3/5`
[∵ `cos^-1x=sin^-1sqrt(1-x^2)]`
`=sin^-1*5/13+sin^-1* 3/5`
`=sin^-1[5/13sqrt(1-(3/5)^2)` `+3/5sqrt(1-(5/13)^2)]`
[∵`sin^1+sin^-1y` `=sin^-1[xsqrt(1-y^2)+ysqrt(1-x^2)]`
`=sin^1[5/13xx4/5+3/5xx12/13]`
`=sin^-1[4/13+36/65]`
`=Sin^-1[56/65]= `R.H.S.
Question: 7 .
`tan^-1*63/16` `=sin^-1*5/13+cos^-1*3/5`
Solution:
R.H.S.`=sin^-1*5/13+cos^-1*3/5` `=sin^-1*5/13+sin^-1sqrt(1-(3/5)^2)`
[∵ `cos^-1x = sin^-1sqrt(1-x^2)]`
`=sin^-1*5/13+sin^-1*4/5`
`=sin^-1[5/13sqrt(1-(4/5)^2)+4/5sqrt(1-(5/13)^2)]`
[∵`sin^-1x+sin^-1y` `=sin^-1[xsqrt(1-y^2+ysqrt(1-x^2)]]`
`sin^-1[5/13xx3/5+4/5xx12/13]`
`=sin^-1[3/13+48/65]`
`=sin^-1[63/65]`
`=tan^-1[(63/65)/sqrt(1-(63/65)^2)]`
[∵`sin^-1x=tan^-1*x/sqrt(1-x^2)]`
`tan^-1[(63/65)/(16/65)]` `tan^-1*63/16=`L.H.S.
Question:8 .
`tan^-1*1/5+tan^-1*1/7+tan^-1*1/3+tan^-1*1/8=pi/4`
Solution:
`tan^-1*1/5+tan^-1*1/7+tan^-1*1/3+tan^-1*1/8`
`(tan^-1*1/3+tan^-1*1/5)` `+(tan^-1*1/7+tan^-1*1/8)`
`tan^-1((1/3+1/5)/(1-1/3xx1/5))` `tan^-1((1/7+1/8)/(1-1/7xx1/8))`
[∵`tan^-1*x+tan^-1y` `=tan^-1(x+y)/(1-xy)]`
`tan^-1(8/14)+tan^-1*(15/55)`
`tan^-1(4/7)+tan^-1*(3/11)`
`tan^-1((4/7+3/11)/(1-4/7xx3/11))`
`tan^-1((44+21)/(77-12))`
`tan^-1*(65/65)`
`=tan^-1(1)=pi/4=`R.H.S.
Reference: