Class 12 Ncert solutions Maths Inverse Trigonometric Functions

Inverse trigonometric function. Exercise 2.2


1. 3sin^-1x=sin^-1(3x - 4x³),   x∈[ -\(\frac{1}{2}\), \(\frac{1}{2}\)]

Sol. Let sin^-1x=y. Then x=sin y. Now

 sin^-1(3x - 4x³)
=sin^-1(3siny-4sin³y) 
=sin^-1(sin 3y)                               ∵(3siny-4sin³y)=sin 3y
=3y=3sin^-1x. Hence proved.


2. 3cos^-1x=cos^-1(4x³ - 3x),   x∈[ -\(\frac{1}{2}\), 1]

Sol. Let cos^-1x=y. Then x=cos y. Now

cos^-1(4x³ - 3x)
=cos^-1(4cos³y - 3cosy)
=cos^-1(cos 3y)                            ∵4cos³y - 3cosy=cos 3y
=3y=3cos^-1x. Hence proved.


3. tan^-1(\(\frac{2}{11}\) )+ tan^-1(\(\frac{7}{24}\) )  =tan^-1(\(\frac{1}{2}\) )   

Sol. By property 5(i) in your textbook (page 44), we have

L.H.S = tan^-1(\(\frac{2}{11}\) )+ tan^-1(\(\frac{7}{24}\) )

= tan^-1(\(\frac{\frac{2}{11}+\frac{2}{24}}{1-\frac{2}{11}\ x \frac{7}{24}}\) )

=tan^-1(\(\frac{\frac{48+77}{11\times24}}{\frac{11\times24-14}{11\times24} }\) )

= tan^-1(\(\frac{\frac{125}{264}}{\frac{250}{264} }\) )

=tan^-1(\(\frac{125}{250} \) )

=tan^-1(\(\frac{1}{2} \) )
=R.H.S  Hence proved.

4. 2tan^-1(\(\frac{1}{2}\) )+ tan^-1(\(\frac{1}{7}\) )  =tan^-1(\(\frac{31}{17}\) )  

Sol. By property 6(iii) (page 44)  2tan^-1x=tan^-1(\(\frac{2x}{1-x^2}\) )

We have 2tan^-1(\(\frac{1}{2}\) )= tan^-1(\(\frac{2\times\frac{1}{2}}{1-\frac{1}{2^2}}\) )

⇒2tan^-1(\(\frac{1}{2}\) )= tan^-1(\(\frac{1}{1-\frac{1}{4}}\) )

⇒2tan^-1(\(\frac{1}{2}\) )= tan^-1(\(\frac{1}{\frac{4-1}{4}}\) )

⇒2tan^-1(\(\frac{1}{2}\) )= tan^-1(\(\frac{4}{3}\) )

 ∴L.H.S= 2tan^-1(\(\frac{1}{2}\) )+ tan^-1(\(\frac{1}{7}\) )

= tan^-1(\(\frac{4}{3}\) ) + tan^-1(\(\frac{1}{7}\) )

=tan^-1(\(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}\ \times \frac{1}{7}}\) )

=tan^-1(\(\frac{\frac{28+3}{3\times7}}{\frac{7\times3-4}{7\times3} }\) )

=tan^-1(\(\frac{\frac{31}{21}}{\frac{17}{21} }\) )

==tan^-1(\(\frac{31}{17} \) )
=R.H.S ..... Hence Proved


Write the following functions in the simplest form: 


5. tan^-1(\(\frac{\sqrt{1+x^2} -1}{x}\) ),   x ≠ 0

Sol.  Let x=tany.   ⇒ y=tan^-1x

 ∴ tan^-1(\(\frac{\sqrt{1+x^2} -1}{x}\) )= tan^-1(\(\frac{\sqrt{1+tan^2 y} -1}{tany}\) )=  tan^-1(\(\frac{\sqrt{sec^2 y} -1}{tany}\) )=tan^-1(\(\frac{sec y -1}{tanyx}\) )
= tan^-1(\(\frac{\frac{1}{cosy} -1}{\frac{siny}{cosy}}\) )=    tan^-1(\(\frac{\frac{1-cosy}{cosy}}{\frac{siny}{cosy}}\) )=     tan^-1(\(\frac{1-cosy}{siny}\) )=  tan^-1(\(\frac{2sin^2 \frac{y}{2}}{2sin \frac{y}{2}cos \frac{y}{2}}\) )=  tan^-1(\(\frac{sin \frac{y}{2}}{ cos \frac{y}{2}}\))  = tan^-1(tan\(\frac{y}{2}\) )=\( \frac{y}{2}\)= \( \frac{1}{2}\)tan^-1x
   

6. tan^-1( \(\frac{1}{\sqrt{x^2  - 1}}\) ) , |x| > 1

Sol. Let x=cosecy.
tan^-1( \(\frac{1}{\sqrt{cosec^2 y - 1}}\) )= tan^-1( \(\frac{1}{\sqrt{cot^2 y}}\) )=tan^-1( \(\frac{1}{coty}\) )=tan^-1( tany )= y =  \(\cosec^{-1} x\)
= \(\frac{\pi}{2}\)-\(\sec^{-1} x\)



7. tan^-1( \(\sqrt{\frac{1-cosx}{1+cosx}}\) ),   x< π 

Sol.  tan^-1( \(\sqrt{\frac{1-cosx}{1+cosx}}\) )= tan^-1( \(\sqrt{\frac{2sin^2 \frac{x}{2}}{2cos^2 \frac{x}{2}}}\) ) = tan^-1( \(\sqrt{\frac{sin^2 \frac{x}{2}}{2cos^2 \frac{x}{2}}}\) ) =tan^-1( \(\sqrt{tan^2 \frac{x}{2}}\) ) = tan^-1( \(tan \frac{x}{2}\) ) = \( \frac{x}{2}\)

8.tan^-1( \(\frac{cosx-sinx}{cosx+sinx}\) ),  0 <x< π 

Sol.  tan^-1( \(\frac{cosx-sinx}{cosx+sinx}\) )=  tan^-1( \(\frac{cosx-sinx}{cosx+sinx}\) ) = tan^-1( \(\frac{1-\frac{sinx}{cosx}}{1+\frac{sinx}{cosx}}\) ) = tan^-1( \(\frac{1-tanx}{1+tanx}\) )
=  tan^-1( \(\frac{tan\frac{\pi}{4}-tan x}{1+tan\frac{\pi}{4}tanx}\) ) =  tan^-1( tan(\(\frac{\pi}{4}\)-x)) = \(\frac{\pi}{4}\)-x

9. tan^-1( \(\frac{x}{\sqrt{a^2 - x^2}}\) ) , |x| < a

Sol.  Let x=asiny.

tan^-1( \(\frac{x}{\sqrt{a^2 - x^2}}\) )=tan^-1( \(\frac{x}{\sqrt{a^2 - asin^2 y}}\) )=tan^-1( \(\frac{x}{\sqrt{a^2(1 - sin^2 y)}}\) )= tan^-1( \(\frac{x}{\sqrt{a^2(cos^2 y)}}\) )= tan^-1( \(\frac{asiny}{acosy}\) )  = tan^-1( tany )  =  y  =\(\sec^{-1} \frac{x}{a}\)

10. tan^-1( \(\frac{3a^2 x - x^3}{a^3 -3ax^2}\) ),   0< a ; \(\frac{-a}{\sqrt{3}} \) ≤ x ≤   \(\frac{a}{\sqrt{3}} \) 

Sol.  Let x=atany.
  tan^-1( \(\frac{3a^2 x - x^3}{a^3 -3ax^2}\) )= tan^-1( \(\frac{3a^2 atany - a^3tan^3y}{a^3 -3aa^2tan^2y}\) )= tan^-1( \(\frac{3a^2 atany - a^3tan^3y}{a^3 -3aa^2tan^2y}\) )
= tan^-1( \(\frac{a^3(3tany - tan^3y)}{a^3(1 -3tan^2y)}\) )=tan^-1( \(\frac{3tany - tan^3y}{1 -3tan^2y}\) ) = tan^-1( tan 3y) = 3y= 3\(\tan^{-1} \frac{x}{a}\)


Find the values of each of the following :

11. tan^-1[2cos(\(\sin^{-1} \frac{1}{2}\))]

Sol. tan^-1[2cos(\(\sin^{-1} \frac{1}{2}\))] =tan^-1[2cos(\( \frac{\pi}{6}\))] = tan^-1[2(\( \frac{1}{2}\))]

= tan^-1[1]= \(\frac{\pi}{4}\)                      ∵\(\sin^{-1} \frac{1}{2}\)=\(\sin^{-1} sin(\frac{\pi}{6})\) =\(\frac{\pi}{6}\) 

  

12. cot  (\(\tan^{-1}a+ cot^{-1}a\))

Sol. cot  (\(\tan^{-1}a+ cot^{-1}a\)) =cot  (\(\pi\over2\))=0 By property 4(ii) page 43

13. tan  \({1\over2}[sin^{-1}{2x\over 1+x^2}+cos^{-1}{1-y^{-2}\over1+y^2}]\),  |x|< 1, y>0 and xy<1 

Sol.  tan  \({1\over2}[sin^{-1}{2x\over 1+x^2}+cos^{-1}{1-y^{-2}\over1+y^2}]\)=tan  \({1\over2}[2tan^{-1}x+2tan^{-1}y]\)

=tan  \([tan^{-1}x+tan^{-1}y]\)

=\({\tan(tan^{1}x)+ tan(tan^{-1}y)\over 1-tan(tan^{-1}x)tan(tan^{-1}y) } \)

=\({ x+y\over1-xy}\)

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