(i) Given as tan^-1(tan π/3)

As tan^-1(tan x) = x if x ϵ [-π/2, π/2]

On, applying this condition in the given question we get,

tan^-1(tan π/3) = π/3

(ii) Given as tan^-1(tan 6π/7)

As we know that tan 6π/7 can be written as (π – π/7)

tan(π – π/7) = – tan π/7

As we know that tan^-1(tan x) = x if x ϵ [-π/2, π/2]

tan^-1(tan 6π/7) = – π/7

(iii) Given as tan^-1(tan 7π/6)

As we know that tan 7π/6 = 1/√3

On, substituting this value in tan^-1(tan 7π/6) we get,

tan^-1 (1/√3)

Let tan^-1 (1/√3) = y

tan y = 1/√3

tan (π/6) = 1/√3

So, the range of the principal value of tan^-1 is (-π/2, π/2) and tan (π/6) = 1/√3

So, tan^-1(tan 7π/6) = π/6

(iv) Given as tan^-1(tan 9π/4)

As we know that tan 9π/4 = 1

On, substituting this value in tan^-1(tan 9π/4) we get,

tan^-1 (1)

Let tan^-1 (1) = y

tan y = 1

tan (π/4) = 1

So, the range of the principal value of tan^-1 is (-π/2, π/2) and tan (π/4) = 1

So, tan^-1(tan 9π/4) = π/4

(v) Given as tan^-1(tan 1)

we have tan^-1(tan x) = x if x ϵ [-π/2, π/2]

On, substituting this condition in given question

tan^-1(tan 1) = 1