1. Find the principal value of sin^-1 (^-1/2)?

Ans : Let sin^-1 (^-1/2) = y

       Then sin y = ^-1/2

                         = - sin (^π/6) = sin (^-π/6 )

      We know that the range of the principal value branch of sin^-1 is

                   [^-π/2 , ^-π/2 ] and sin [^-π/6 ] = ^-1/2

                   ∴  the principal value of sin-1 (^-1/2) is ^{- π}/6.

2. Find the principal value of cos^-1 (^√3/2)?

Ans : Let cos^-1 (^√3/2) = y.

         Then , cos y = ^√3/2 = cos (^π/6 ).

          We know that the range of the principal value branch of cos^-1 is [ 0 , π ] and cos (^π/6 ) = ^√3/2.

     Therefore, the principal value of cos^-1 (^√3/2) is ^π/6.

3. Find the principal value of tan^-1 (-1)?

Ans : Let tan^-1 (-1) = y

         Then , tan y = -1 = - tan (^π/4) = tan (^{- π}/4 )

         We know that the range of the principal value branch of tan^-1 is

             [ ^-π/2 , ^π/2 ] and tan (^-π/4) = -1

         Therefore, the principal value of tan^-1(-1) is ^-π/4

4. Find the value of cos^-1(¹/2) + 2 sin^-1 (¹/2) ?

Ans : Let cos^-1(¹/2) = x

         Then , cos x = ¹/2 = cos (^π/3)

          Cos^-1 (¹/2) = ^π/3

        Let sin^-1 (¹/2) = y. Then sin y = ¹/2 = sin (^π/6).

          Sin^-1 (¹/2) = ^π/6

          cos^-1(¹/2) + 2 sin^-1 (¹/2) = ^π/3 + ^2π/6 = ^π/3 + ^π/3 = ^2π/3

5. Prove 3 sin-1 x = sin-1 (3x - 4x³), x ∈ [^-1/2, ¹/2 ]?

Ans : To prove 3 sin^-1 x = sin^-1 (3x - 4x³), x ∈ [^-1/2, ¹/2 ]

          Let x = sin 6

          Then , sin^-1 x = 6

          We have,

            R.H.S = sin^-1 (3x - 4x³ )

                      = sin^-1 (3 sin6 - 4 sin36 )

                      = sin^-1 (sin 36)

                      = 36

                      = 3 sin^-1 (x)

                      = L .H .S

6. Prove tan^{-1 2}/11 + tan^{-1 7}/24 = tan^{-1 1}/2?

Ans : L.H .S = tan^{-1 2}/11 + tan^{-1 7}/24

7. Write the function in the simplest form :

Ans : 

8. Write the function in the simplest form :

Ans :

9. Find the value of cos-1 (cos ^13π/0 ) ?

Ans : We know that cos^-1(cos x ) = x if x∈ [ 0 , π ] , which is the principal value branch of cos^-1x.

          Here, ¹³/6 ∉  [ 0 , π ]

          Now, cos^-1 (cos ^13π/6 ) can be written as

                cos^-1 (cos ^13π/6 ) = cos^-1 [ cos (2 π + ^π/6 ) ]

                                              = cos^-1 [ cos ( ^π/6) ], where ^π/6 ∈ [ 0 , π ]

           ∴    cos^-1 (cos ^13π/6 ) = cos^-1 [ cos ( ^π/6) ] = ^π/6

10. Prove that 2 sin^{-1 3}/5 = tan^{-1 24}/7?

Ans : Let sin^{-1 3}/5 = x

         Then sinx = ³/5

       ∴  tanx = ³/4

       ∴  x = tan^{-1 3}/4

            ⇒  sin^{-1 3}/5 = tan^{-1 3}/4

       Now , we have:

        L.H.S = 2 sin^{-1 3}/5 = 2 tan^{-1 3}/4