Question

(a) sin2+ ______=1. (b) sin2=12 ∴=_____

Answer 1

Option B is correct.

Step-by-step explanation:

Given that cos(a+b)=0

then we have to find the reduced form of sin(a-b)

As, cos(a+b)=0

⇒ a+b=cos^{-1}(0)=90^{\circ}

⇒ a=(90-b)^{\circ}

sin(a-b)=sin(90-b-b)=sin(90-2b)=cos{2b}

Hence, sin(a-b)=cos{2b}

∴ option b) is correct.

Answer 2

Answer:

-1

14

Step-by-step explanation: