LHS=2t^2/1-t^2
RHS=2t^2/1-t^2
Hence left had side is equal
Step-by-step explanation:
Taking tan x and sec x in the terms of t=tan(x/2)
tan x=2t/1-t^2
sex x=1/cos x
therefore cos x=1-t^2/1+t^2
so sec x=1/1-t^2/1+t^2
sec x=1+t^2/1-t^2
Taking tan x and sec x in the terms of t=tan(x/2)
tan x=2t/1-t^2
sex x=1/cos x
therefore cos x=1-t^2/1+t^2
so sec x=1/1-t^2/1+t^2
sec x=1+t^2/1-t^2
LHS=tanx .tan(x/2)
=2t.t/1-t^2
=2t^2/1-t^2
RHS=sec x-1
=1+t^2/1-t^2-1
=1+t^2-1+t^2/1-t^2
=2t^2/1-t^2
Therefore hence we proved LHS=RHS
=2t.t/1-t^2
=2t^2/1-t^2
RHS=sec x-1
=1+t^2/1-t^2-1
=1+t^2-1+t^2/1-t^2
=2t^2/1-t^2
Therefore hence we proved LHS=RHS
