Question

If sin2 x + cos2 y = 2 sec2 z, then general solution of triplets (x, y, z) is

Answer 1

x=(n+12)π, y=mπ∴x=n+12π, y=mπ and z = rπ where n∈I, m∈I, r∈I

Explanation:

∴ LHS ≤ 2 and RHS ≥ 2

So, sin2 x = 1, cos2 y = 1 and sec2 z = 1

∴x=(n+12)π, y=mπ∴x=n+12π, y=mπ and z = rπ where n∈I, m∈I, r∈I