please answer the question
Attachments:

Answers
Answered by
0
Answer:
Step-by-step explanation:
jpatar16:
yes
Answered by
0
Hii.....
sin(A+B) = sinA. cosB + cosA. sinB
cos(A-B) = cosA . cos B + sinA . sinB
{cos(90-a)=sina}
{sin(90-a)=cosa}
cos(90-A-B)=cos(90-A).cos B + sin(90-A).sinB
cos(90-(A+B)= sinA.cosB + cosA.sinB
sin(A+B)=sinA . cosB + cosA . sinB.
sin(A+B) = sinA. cosB + cosA. sinB
cos(A-B) = cosA . cos B + sinA . sinB
{cos(90-a)=sina}
{sin(90-a)=cosa}
cos(90-A-B)=cos(90-A).cos B + sin(90-A).sinB
cos(90-(A+B)= sinA.cosB + cosA.sinB
sin(A+B)=sinA . cosB + cosA . sinB.
Similar questions
Math,
9 months ago
English,
9 months ago
Geography,
9 months ago
Social Sciences,
1 year ago
Chemistry,
1 year ago