Question

pls answer the following question

Answer 1

heya brother....✌

Here is your answer...
==================
From given...

cos(@+B)=0

cos(@+B)=cos90°【As you know that. cos90°=0】

so,

=>@+B=90°

=>@=90°-B-------1)

then......⏩

⏩sin(@-B)=?

so,sin[(90°-B)-B)].【from equation 1]

=>sin(90°-2B)

=>Hence 【angle is in 1st quadrant 】

so,.

cos2B is correct answer..【.°.sin(90°-¢)=cos¢】

now I am sure ur answer is 2nd option cos2B...

hope it help you....

@Rajukumar☺☺☺......

if it is not correct then plz report it..