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# Class 12-science NCERT Solutions Maths Chapter 2 - Inverse Trigonometric Functions

## Inverse Trigonometric Functions Exercise Ex. 2.1

### Solution 1 ### Solution 2 ### Solution 3 ### Solution 4 ### Solution 5 ### Solution 6 ### Solution 7 We know that the range of the principal value of sec-1 is ### Solution 8 ### Solution 9 ### Solution 10 ### Solution 11 ### Solution 12 ### Solution 13 ### Solution 14  ## Inverse Trigonometric Functions Exercise Ex. 2.2

### Solution 1  ### Solution 2 ### Solution 3 ### Solution 4 ### Solution 5 ### Solution 6 ### Solution 7 ### Solution 8 ### Solution 9 ### Solution 10 ### Solution 11 ### Solution 12 ### Solution 13 ### Solution 14

sin (sin-1 (1/5) + cos-1x) = 1

∴ sin-1 (1/5) + cos-1 x = sin-1 1

sin-1 (1/5) + cos-1 x = π/2

sin-1 (1/5) = π/2 - cos-1 x

sin-1 (1/5) = sin-1 x

x = 1/5

We use the results: sin-1 1 = π/2 and sin-1 x + cos-1 x = π/2

Concept Insight:

As R.H.S is ‘1’, it is easier to take sin-1 1 = π/2

### Solution 15 ### Solution 16 ### Solution 17 ### Solution 18  ### Solution 19 ### Solution 20 ### Solution 21 ## Inverse Trigonometric Functions Exercise Misc. Ex.

### Solution 1 ### Solution 2 ### Solution 3 ### Solution 4 ### Solution 5  = by (3)

= RHS ### Solution 6

Let sin-1(3/5) = A and cos-1 (12/13) = B

So sin A = 3/5 and cos B = 12/13

Hence cos A = 4/5 and sin B = 5/13

As R.H.S is sin-1 we use sin (A + B)

Sin (A + B) = sin A cos B + cos A sin B = (3/5) (12/13) + (4/5) (5/13)

= 36/65 + 20/65 = 56/65

Thus A + B = sin-1 (56/65) hence proved.

Concept insight:

If R.H.S is cos-1 or sin-1 then use Cos (A + B) or sin (A + B) as the case may be.

### Solution 7 ### Solution 8 ### Solution 9 ### Solution 10 ### Solution 11 ### Solution 12 ### Solution 13 ### Solution 14 ### Solution 15 ### Solution 16  ### Solution 17 