What is the value of sin(48° + θ) - cos(42° - θ)?

Correct option is D

Given:

sin(48° + θ) - cos(42° -  θ)

Formula used:

$$\sin x = \cos(90^\circ - x)$$​

Solution:

x = 48° +θ  

From complementary angle theorem;

sin(48° + θ) = cos(90°- 48°- θ)

sin(48° +θ) = cos( 42°- θ) 

So, 

sin(48° + θ) - cos( 42°- θ) = 0