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**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 - 4x ^{3}), x ∈ [^{-1}/2, ^{1}/2 ]?**

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

Let x = sin 6

Then , sin^{-1} x = 6

We have,

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

= 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^{-1}x.

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

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Plus 2 Computer Science

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