Question

Simplify $(cos((x-y)/2)/cos(x))-tan(x)*sin((x+y)/2)$ Answer urgently needed

Answer 1

Answer:

Step-by-step explanation:

Cos(x-y/2) / Cosx - Tanx.Sin(x+y/2)

=> Cos (x-y/2)/Cosx - Sinx/Cosx * Sin(x + y/2)

//1/Cosx in denominator is common

=> 1/Cosx[ Cos(x-y/2) - Sinx*Sin(x+y/2)]

//Sinx can be written as Sin (x+y+x - y/2) = Sin(x +y/2 + x - y/2)

=> 1/Cosx[ Cos(x-y/2) - Sin(x+y/2 + x-y/2).Sin(x+y/2)]

//We know that Sin(A+B) = SinA CosB + CosASinB

=> 1/Cosx [Cos(x-y/2) - (Sin(x+y/2)Cos(x-y/2) + Cos(x+y/2)Sin(x-y/2))Sin(x+y/2)]

=> 1/Cosx[Cos(x-y/2) - Sin²(x+y/2)(Cosx-y/2) - Cos(x+y/2)Sin(x-y/2)Sin(x+y/2)]

=>1/Cosx[ Cos(x-y/2){1 - Sin²(x+y/2)} - Cos(x+y/2)Sin(x-y/2)Sin(x+y/2)]

// We know that 1 - Sin²θ = Cos²θ

=> 1/Cosx[Cos(x-y/2).Cos²(x+y/2) - Cos(x+y/2)Sin(x-y/2)Sin(x+y/2)]

//Take Cos(x+y/2) common

=>Cos(x+y/2)/Cosx[Cos(x-y/2).Cos(x+y/2) - Sin(x-y/2)Sin(x+y/2)]

//The expression in brackets is of form CosACosB - SInASinB which is equal to Cos (A + B)

=> Cos(x+y/2)/Cosx [ Cos(x+y/2 + x - y/2)]

=> Cos(x+y/2)/Cosx [ Cos2x/2]

=> Cos(x+y/2)/Cosx * Cosx

//Cosx in numerator and deominator strike out.

=> Cos(x+y/2)