Question

In triangle abc,b^2cos2A-a^2cos2B=

Answer 1

b^2(2cos^2A-1)-a^2(2cos^2B-1)
=>( 2b^2cos^2A-2a^2cos^2B)+(a^2-b^2)
=>2(bcosA-acosB)(bcosA+acosB)+(a^2-b^2)
you know acosB+bcosA=c
use this
=> 2c(bcosA-acosB)+(a^2-b^2)
now use cosine formula formula A and B
=>(b^2+c^2-a^2-c^2-a^2+b^2)+(a^2-b^2)
=> b^2-a^2

Answer 2

Answer:

Your answer is b² - a²

Step-by-step explanation:

Explanation is given in the attachment.......

Happy studying...