Question -
Answer -
(i) Given tan-1(tanπ/3)
As tan-1(tanx) = x if x ϵ [-π/2, π/2]
By applying thiscondition in the given question we get,
Tan-1(tanπ/3) = π/3
(ii) Given tan-1(tan6π/7)
We know that tan 6π/7can be written as (π – π/7)
Tan (π – π/7) = – tanπ/7
We know that tan-1(tanx) = x if x ϵ [-π/2, π/2]
Tan-1(tan6π/7) = – π/7
(iii) Given tan-1(tan7π/6)
We know that tan 7π/6= 1/√3
By substituting thisvalue in tan-1(tan 7π/6) we get,
Tan-1 (1/√3)
Now let tan-1 (1/√3)= y
Tan y = 1/√3
Tan (π/6) = 1/√3
The range of theprincipal value of tan-1 is (-π/2, π/2) and tan (π/6) = 1/√3
Therefore tan-1(tan7π/6) = π/6
(iv) Given tan-1(tan9π/4)
We know that tan 9π/4= 1
By substituting thisvalue in tan-1(tan 9π/4) we get,
Tan-1 (1)
Now let tan-1 (1)= y
Tan y = 1
Tan (π/4) = 1
The range of theprincipal value of tan-1 is (-π/2, π/2) and tan (π/4) = 1
Therefore tan-1(tan9π/4) = π/4
(v) Given tan-1(tan1)
But we have tan-1(tanx) = x if x ϵ [-π/2, π/2]
By substituting thiscondition in given question
Tan-1(tan 1)= 1
(vi) Given tan-1(tan2)
As tan-1(tanx) = x if x ϵ [-π/2, π/2]
But here x = 2 whichdoes not belongs to above range
We also have tan (π –θ) = –tan (θ)
Therefore tan (θ – π)= tan (θ)
Tan (2 – π) = tan (2)
Now 2 – π is in thegiven range
Hence tan–1 (tan2) = 2 – π
(vii) Given tan-1(tan4)
As tan-1(tanx) = x if x ϵ [-π/2, π/2]
But here x = 4 whichdoes not belongs to above range
We also have tan (π –θ) = –tan (θ)
Therefore tan (θ – π)= tan (θ)
Tan (4 – π) = tan (4)
Now 4 – π is in thegiven range
Hence tan–1 (tan2) = 4 – π
(viii) Given tan-1(tan12)
As tan-1(tanx) = x if x ϵ [-π/2, π/2]
But here x = 12 whichdoes not belongs to above range
We know that tan (2nπ– θ) = –tan (θ)
Tan (θ – 2nπ) = tan(θ)
Here n = 2
Tan (12 – 4π) = tan(12)
Now 12 – 4π is in thegiven range
∴ tan–1 (tan12) = 12 – 4π.