Question

A+B+C=90 then Sin2A+sin2B- sin2C

Answer 1

Answer:

4sinAsinBcosC

Step-by-step explanation:

Hi,

Given that A + B + C = 90,

Sin2A+sin2B- sin2C

= ( sin2A + sin2B ) - sin2C

= 2sin(A + B)cos(A - B) - sin2C

But  A+ B = 90 - C

= 2sin(90 - C)cos(A - B) - sin2C

= 2cosC.cos(A - B) - 2sinCcosC

= 2cosC[cos(A - B) - sinC]

But C = 90 - A - B

= 2cosC[cos(A - B) - sin(90 - A - B)]

= 2cosC[cos(A - B) - cos( A + B)]

= 2cosC[2sinAsinB]

= 4sinAsinBcosC

Hope, it helps !

Answer 2

Step-by-step explanation:

I hope it is helpful to you ☺️☺️