Question

In triangle ABC if angleC=pi/2 then prove that sin(A-B)=(a^2-b^2)/a^2+b^2​

Answer 1

Hey !!!

from question

a²+b²/a²-b² =sin(A+B)/sin(A-B)

(by using componendo and dividendo rule )

a²+b²+a²-b²/a²+b²-a²+b² = sin(A+B)+sin(A-B)/sin(A+B) - sin(A-B)

or, a²/b² =. 2sinA×cosB/2cosA×sinB

or,

ksin²A/ksin²B = 2sinA×cosB/2cosA×sinB

or, sinA/sinB =cosB/cosA

or, sinA×cosA =cosA×sinB {° • ° sinA=/ 0, sinB=/0}

or, 2sinA×cosA =2sinB×cosB

or, sin2A - sin2B =0

or, 2cos(A+B) sin(A-B) =0

•°• cos(A+B ) =0 or sin(A-B ) =0

when , cos(A+B) =0

.then

cos(A+B) =cos90°

A+B = 90°

•°• C =(180° - (A+B ) =180°-90° =90° -----------1)

when,

sin(A-B ) = 0

then, A-B =0

•°• A =B--------------2)

from 1) and 2 ) triangle either right angled triangle or iscosecles triangle .

note ➡sin2A = sin2B => 2A =2B

or, 2A =180° -2B

=> A = B

or, A+B =90°

or , c =90°

hope it helps !!!

#Shaun Abraham