1) Define inverse - Sine function.

                   The iverse - sine function is defined as: y = f^-1(x) = sin^-1x iff x = siny and

The graph of sin^-1x is 

Domain of sin^-1x = [-1, 1] and Range of sin^-1 x =

2) Define inverse-cosine function. 

             The inverse - cosine function is defined as y = f^-1(x) = cos^-1x iff x = cosy and

Domain of cos^-1x = [-1, 1] and 

Range of cos^-1x =

The graph of cos^-1x is 

3) Define inverse-tangent function. 

           The inverse tangent function is defined as y = f^-1(x) = tan^-1x iff x = tan y and 

The graph of tan^-1 x is 

Domain of tan^-1x = R and 

Range of tan^-1x =

4) Define inverse-cotangent function.

             The inverse cotangent function is defined as y = f^-1(x) = cot^-1x iff x = cot y and

The graph of cot^-1x is 

Domain of cot^-1x = R 

and Range of cot^-1x =

5) Define inverse - secant function.

             The inverse secant function is defined as y = f^-1(x) = sec^-1x iff x = sec y and

The graph of sec^-1x is 

Domain of sec^-1x =

Range of sec^-1x =

6) Define inverse-cosecant function. 

            The inverse cosecant function is defined as y = f^-1(x) = cosec^-1 x iff 

x = cosec y and

Domain of cosec-1x =

Range of cosec^-1x =

The graph of cosec^-1 x is

7. Find the principal values of sin^-1 (-1) and cot^-1 (√3) 

8. Write principal value of cos^-1(cos 5π/3) 

9. What is the principal value of cos^-1 (cos 2π/3) + sin^-1 (sin 2π/3)? 

10. Find the range of one branch of sin^-1x, other than the principal branch:-

By definition; 

-----, -3π/2 ≤ sin^-1 x ≤ -π/2

-π/2≤ sin^-1 x ≤ π/2, π/2 ≤ sin^-1 x ≤ 3π/2,........

Hence one branch, other than principal branch is π/2 ≤ sin^-1 x ≤ 3π/2. 

11. Find three branches, other than the principal value branch, of tan^-1x. 

Let y = tan^-1x

By definition, the inverse function of tanx exists in each of the intervals 

-3π/2<y<-π/2 ,π/2<y<3π/2 ,3π/2 <y<5π/2 ,etc.

Hence the required three branches of tan^-1x, other than the principal value branch are: 

-3π/2 < tan^-1x < -π/2, π/2 < tan^-1x <3π/2

and 3π/2 <tan^-1x < 5π/2.

12. Find the value of sin^-1 (sin 3π/5)  

Since sin^-1(sinx) = x, 

 sin^-1 (sin 3π/5) = 3π/5

But , which is the principal branch of sin^-1x

And sin 3π/5 = sin (π - 3π/5) = sin 2π/5 

13. Find the value of tan^-1 [2 cos (2 sin^-1 1/2)]

Put 2sin^-1 1/2 = t 

implies sin^-1 1/2 = t/2

implies 1/2 = sin t/2 

implies  sin t/2 = 1/2------------ (1) 

Now tan^-1 [2 cos (2sin^-1 1/2)] = tan^-1 [2 cos t] 

= tan^-1 [2 (1 - 2 sin² t/2)] 

= tan^-1[2 (1 - 2(1/4))] 

= tan^-1 [2(1 - 1/2)]

= tan^-1 [2 (1/2)]

= tan^-1 (1)

= π/4

14. Find the value of tan (sin^-1 3/5 + cot^-1 3/2) 

15. Prove that tan^-1(1/2) + tan^-1 (1/5) + tan^-1(1/8) = π/4 

16. Show that sin^-1 3/5 - sin^-1 8/17 = cos^-1 84/85 

Let sin^-1 3/5 = x and sin^-1 8/17 = y 

implies sinx = 3/5 and sin y = 8/17 

17. Solve the following for x 

tan^-1  = π/4 + tan^-1x, 0 < x < 1

This is true for all x ε (0,1) 

Hence the solution set is (0, 1) 

18. Solve sin(sin^-1 1/5 + cos^-1 x) = 1

We have: sin (sin^-1 1/5 + cos^-1x) = 1 = sin π/2 

implies sin^-1 1/5 + cos^-1 x = π/2

ie,  cos^-1x = π/2 - sin^-1 1/5 

                = cos^-1 1/5

Hence x = 1/5

19. Solve tan^-1x + 2 cot^-1x = 2π/3

We have tan^-1x + 2 cot^-1x = 2π/3 

implies  (tan^-1x + cot^-1x) + cot^-1x = 2π/3

implies π/2 + cot^-1x = 2π/3

ie , cot^-1x = 2π/6 - π/2

implies cot^-1x = π/6 

ie,  x = cot π/6 

ie, x = √3

20. Write  in the simplest form.