Question

It A+B+C=180° then show that sin2a-sin2B+
sin2c=4 COSA SinB cosc.​

Answer 1

Answer:

D

Step-by-step explanation:

• sin2A-sin2B+sin2C
• 2cos(180-C)sin(A-B)+sin2C
• -2cosCsin (A-B)+2sinCcosC
• 2cosC[-sin(A-B)+sin(180-A-B)]
• 2cosC-sin(A-B)+sin(A+B)
• 2cosC×2cosAsinB
• 4cosAsinBcosC
• hence, option D is correct