If sin(3x - 20$$)^o$$ = cos (20 - 3y$$)^o$$, then value of x - y will be :​

Correct option is C

Given: 

We are given the equation:

​$$\sin(3x - 20^\circ) = \cos(20^\circ - 3y)$$​​

Concept Used:

​$$\cos(\theta) = \sin(90^\circ - \theta)$$  

Solution:

Applying the identity to the right-hand side:

​$$\cos(20^\circ - 3y) = \sin(90^\circ - (20^\circ - 3y)) = \sin(70^\circ + 3y)$$ 

Now,

$$\sin(3x - 20^\circ) = \sin(70^\circ + 3y)$$​

$$3x - 20^\circ = 70^\circ + 3y$$ 

$$3x - 3y = 90^\circ$$ 

$$x - y = 30^\circ$$ 

​Thus, the value of $x-y$ is: $$30^\circ$$​​

180∘180^\circ Therefore, we have two cases to consider:3x−20∘=70∘+3y3x - 20^\circ = 70^\circ + 3y3x−20∘=70∘+3y