Question -
Answer -
Let us consider LHS:
(sec x sec y + tan x tan y)2┬атАУ (sec xtan y + tan x sec y)2
Expanding the above equation we get,
[(sec x secy)2┬а+ (tan x tan y)2┬а+ 2 (sec x sec y) (tan xtan y)] тАУ [(sec x tan y)2┬а+ (tan x sec y)2┬а+ 2(sec x tan y) (tan x sec y)] [sec2┬аx sec2┬аy +tan2┬аx tan2┬аy + 2 (sec x sec y) (tan x tan y)]тАУ [sec2┬аx tan2┬аy + tan2┬аx sec2┬аy+ 2 (sec2┬аx tan2┬аy) (tan x sec y)]
sec2┬аx sec2┬аy тАУ sec2┬аxtan2┬аy + tan2┬аx tan2┬аy тАУ tan2┬аxsec2┬аy
sec2┬аx (sec2┬аy тАУ tan2┬аy)+ tan2┬аx (tan2┬аy тАУ sec2┬аy)
sec2┬аx (sec2┬аy тАУ tan2┬аy)тАУ tan2┬аx (sec2┬аy тАУ tan2┬аy)
We know, sec2┬аx тАУ tan2┬аx= 1.
sec2┬аx ├Ч 1 тАУ tan2┬аx ├Ч1
sec2┬аx тАУ tan2┬аx
1 = RHS
тИ┤ LHS = RHS
Hence proved.