Question -
Answer -
Let us consider LHS:
(sec x sec y + tan x tan y)2 тАУ (sec xtan y + tan x sec y)2
Expanding the above equation we get,
[(sec x secy)2 + (tan x tan y)2 + 2 (sec x sec y) (tan xtan y)] тАУ [(sec x tan y)2 + (tan x sec y)2 + 2(sec x tan y) (tan x sec y)] [sec2 x sec2 y +tan2 x tan2 y + 2 (sec x sec y) (tan x tan y)]тАУ [sec2 x tan2 y + tan2 x sec2 y+ 2 (sec2 x tan2 y) (tan x sec y)]
sec2 x sec2 y тАУ sec2 xtan2 y + tan2 x tan2 y тАУ tan2 xsec2 y
sec2 x (sec2 y тАУ tan2 y)+ tan2 x (tan2 y тАУ sec2 y)
sec2 x (sec2 y тАУ tan2 y)тАУ tan2 x (sec2 y тАУ tan2 y)
We know, sec2 x тАУ tan2 x= 1.
sec2 x ├Ч 1 тАУ tan2 x ├Ч1
sec2 x тАУ tan2 x
1 = RHS
тИ┤ LHS = RHS
Hence proved.