If x sin 3 θ + y cos 3 θ = sin θ · cos θ and x sin θ - y cos θ = 0 , prove that x 2 + y 2 = 1 .

x sin 3 θ + y cos 3 θ = sin θ cos θ . .. ( 1 ) and x sin θ − y cos θ = 0 . .. ( 2 ) From 2 x sin θ = y cos θ Multiple both sides by sin 2 θ x sin 3 θ = y cos θ sin 2 θ Put in 1 y cos θ sin 2 θ + y cos 3 θ = sin θ cos θ y cos θ ( sin 2 θ + cos 2 θ ) = sin θ . cos θ , using sin 2 θ + cos 2 θ = 1 y = sin θ Similarly x = cos θ Square and add to get the result. x 2 + y 2 = 1

If x sin 3 θ + y cos 3 θ = sin θ · cos θ and x sin θ - y cos θ = 0 , prove that x 2 + y 2 = 1 .

x sin 3 θ + y cos 3 θ = sin θ cos θ . .. ( 1 ) and x sin θ − y cos θ = 0 . .. ( 2 ) From 2 x sin θ = y cos θ Multiple both sides by sin 2 θ x sin 3 θ = y cos θ sin 2 θ Put in 1 y cos θ sin 2 θ + y cos 3 θ = sin θ cos θ y cos θ ( sin 2 θ + cos 2 θ ) = sin θ . cos θ , using sin 2 θ + cos 2 θ = 1 y = sin θ Similarly x = cos θ Square and add to get the result. x 2 + y 2 = 1

If tan θ = a b , then show that a sin θ - b cos θ a sin θ + b cos θ = a 2 - b 2 a 2 + b 2 .

Given that, tan θ = a b . We have to prove that, a sin θ - b cos θ a sin θ + b cos θ = a 2 - b 2 a 2 + b 2 LHS = a sin θ - b cos θ a sin θ + b cos θ = a sin θ cos θ - b cos θ cos θ a sin θ cos θ + b cos θ cos θ [Divide both the numerator and the denominator by cos θ ] = a tan θ - b a tan θ + b = a × a b - b a × a b + b [Substitute the value of tan θ = a b that is given] = a 2 - b 2 a a 2 + b 2 b = a 2 - b 2 a 2 + b 2 = RHS [Proved]