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# If tan (x+y) = a+b and tan (x-y) = a-b Then prove that a tan x - b tan y = a^2 - b^2

somi teez
105 Points
4 years ago
hi,
first we will try to bring it in form a^2-b^2.
tan (x+y) tan (x-y) = (a+b)*(a-b)
using formula,=(tan^2x-tan^2y)/(1-tan^2x*tan^2y)=a^2-b^2................(1)
now,
tan (x+y)+tan(x-y)=a+b+a-b=2a.....................................................(2)
=(tan x+tan y)/(1-tan x*tan y) + (tan x-tan y)/(1+tan x*tan y)=2a
taking l.c.m and then evaluating,
=2(tan x+tan x*tan^2y)/(1-tan^2x*tan^2y)=2a
(2 cancels on both sides)
and multiplying both sides by tan x we get,
a tan x=(tan^2 x+tan^2 x*tan^2y)/(1-tan^2x*tan^2y)........................(3)
now,   tan (x+y)-tan(x-y)=a+b-a+b=2b
then as above proceeding we get ,
b tan y=(tan^2 y+tan^2 x*tan^2y)/(1-tan^2x*tan^2y)........................(4)
evaluating a tan x - b tan y =(tan^2x-tan^2y)/(1-tan^2x*tan^2y)=a^2-b^2.....(using eqation 1)
HENCE,PROVED...