sin (45° + θ) – cos (45° – θ) equals :(A) 2 cos θ(B) 0(C) 2 s

1.	sin (45° + θ) – cos (45° – θ) equals :(A) 2 cos θ(B) 0(C) 2 sinθ(D) 1
Answer» Answer is (B) 0 sin (45° + θ) – cos (45° – θ) = sin (45° + θ) – cos [90° – (45° + θ)] [∵ (45° – θ) = {90° – (45° + θ)}] = sin (45° + θ) – sin (45° + θ) [∵ cos (90° – θ) = sin θ] = 0

sin (45° + θ) – cos (45° – θ) equals :(A) 2 cos θ(B) 0(C) 2 sinθ(D) 1

Answer»

Answer is (B) 0

sin (45° + θ) – cos (45° – θ)

= sin (45° + θ) – cos [90° – (45° + θ)] [∵ (45° – θ) = {90° – (45° + θ)}]

= sin (45° + θ) – sin (45° + θ) [∵ cos (90° – θ) = sin θ]

= 0