\frac { \operatorname { cos } 4 x + \operatorname { cos } 3 x + \opera

1.	\frac { \operatorname { cos } 4 x + \operatorname { cos } 3 x + \operatorname { cos } 2 x } { \operatorname { sin } 4 x + \operatorname { sin } 3 x + \operatorname { sin } 2 x } = \operatorname { cot } 3 x
Answer» We have to prove (Cos4x+Cos3x+Cos2x)/(Sin4x+Sin3x+Sin2x) =cot3x. So, Using the formula for the first and third terms of numerator : (CosA+CosB)=Cos((A+B)/2)Cos((A-B)/2). =Cos((4x+2x)/2)Cos((4–2)/2) =Cos3xCosx Using the formula for the first and third terms of denominator : (SinA+SinB)=Sin((A+B)/2)Cos((A-B)/2). =Sin((4+2)/2)Cos((4–2)/2) =Sin3xCosx So, Left hand side of given expression becomes =(Cos3x + Cos3xCosx)/(Sin3x + Sin3xCosx) =Cos3x(1+Cosx)/Sin3x(1+Cosx) =cot3x =Right hand side of the given expression.

\frac { \operatorname { cos } 4 x + \operatorname { cos } 3 x + \operatorname { cos } 2 x } { \operatorname { sin } 4 x + \operatorname { sin } 3 x + \operatorname { sin } 2 x } = \operatorname { cot } 3 x

Answer»

We have to prove (Cos4x+Cos3x+Cos2x)/(Sin4x+Sin3x+Sin2x) =cot3x.

So, Using the formula for the first and third terms of numerator :

(CosA+CosB)=Cos((A+B)/2)*Cos((A-B)/2).

=Cos((4x+2x)/2)*Cos((4–2)/2)

=Cos3x*Cosx

Using the formula for the first and third terms of denominator :

(SinA+SinB)=Sin((A+B)/2)*Cos((A-B)/2).

=Sin((4+2)/2)*Cos((4–2)/2)

=Sin3x*Cosx

So, Left hand side of given expression becomes

=(Cos3x + Cos3x*Cosx)/(Sin3x + Sin3x*Cosx)

=Cos3x(1+Cosx)/Sin3x(1+Cosx)

=cot3x =Right hand side of the given expression.