We are given that,

Take,

We can write as,

Then,

Also, by trigonometric identity

[∵, lies in III Quadrant and tangent is positive in III Quadrant]

Using the property of inverse trigonometry, that is, tan^-1(tan A) = A.

Now, take

We can write as,

Then,

By trigonometric identity,

[∵, lies in II Quadrant and tangent is negative in II Quadrant]

Using the property of inverse trigonometry, that is, tan^-1(tan A) = A.

We have,

⇒ 4α = π and 3β = π

Since, the values of 4α and 3β are same, that is,

4α = 3β = π

Therefore,

4α = 3β