Question

If tan(x+y) +tan(x-y) =1,find dy/dx
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Answer 1

Answer:

tan (x+ Y ) + tan(x- y) =0

= sex^2( x+ y) * ( 1 + dy/dx) + sec^2( x-y)*( 1- dy/dx) =0

= sec^2(x+y) + sec^2(x-y ) + dy/dx[ sec^2(x+y) - sec^2(x-y) ]=0

dy/dx = -{ sec^2(x+ y) + sec^2 (x- y) }/{sec^2(x+y)-sec^2(x-y)}

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Answer 2

dy/dx = sec^(x+y) [1+dy/dx] + sec^(x-y) [1-dy/dx]

dy/dx = sec^2(x+y) + sec^2(x-y) + dy/dx[sec^2(x+y) - sec^2(x-y)]

bring dy/dx to LHS

dy/dx[1+sec^2(x-y)-sec^2(x+y)] = sec^2(x+y) + sec^2(x-y)

or dy/dx = (sec^2(x+y) + sec^2(x-y)) / (1+sec^2(x-y)-sec^2(x+y))