Question

Prove that `cot (A/2)- 3cot ((3A)/2) = (4sinA)/(1+2cosA)`?

Answer 1

Please refer to the Explanation.

Explanation:

We know that,

`tan3theta=(3tantheta-tan^3theta)/(1-3tan^2theta)`.

`:. cot3theta=1/(tan3theta)=(1-3tan^2theta)/(3tantheta-tan^3theta)`

`:.cot((3A)/2)={1-3tan^2(A/2)}/{3tan(A/2)-tan^3(A/2)}`.

Letting `tan(A/2)=t,` we have,

`cot(A/2)-3cot((3A)/2)`,

`=1/t-3{(1-3t^2)/(3t-t^3)}`,

`1/t-{3(1-3t^2)}/{t(3-t^2)}`,

`={(3-t^2)-3(1-3t^2)}/{t(3-t^2)}`,

`=(8t^cancel(2))/{cancel(t)(3-t^2)}`,

`=(8t)/{(1+t^2)+2(1-t^2)}`

`={4*(2t)/(1+t^2)}/{(1+t^2)/(1+t^2)+2*(1-t^2)/(1+t^2)}`.

Note that, `(2t)/(1+t^2)={2tan(A/2)}/(1+tan^2(A/2))=sinA, and`

`(1-t^2)/(1+t^2)=cosA`.

`rArrcot(A/2)-3cot((3A)/2)=(4sinA)/(1+2cosA)," as desired!"`

Answer 2

Please see below.

Explanation:

`LHS=cot(x/2)-3cot((3x)/2)`

`=cos(x/2)/sin(x/2)-3*cos((3x)/2)/sin((3x)/2)`

`=(sin((3x)/2)*cos(x/2)-3*cos((3x)/2)*sin(x/2))/(sin(x/2)*sin((3x)/2)`

`=(2sin((3x)/2)*cos(x/2)-3*2cos((3x)/2)*sin(x/2))/(2sin(x/2)*sin((3x)/2)`

`=(sin((3x)/2+x/2)+sin((3x)/2-x/2)-3*{sin((3x)/2+x/2)-sin((3x)/2-x/2)})/(cos((3x)/2-x/2)-cos((3x)/2+x/2)`

`=(sin((4x)/2)+sin((2x)/2)-3*{sin((4x)/2)-sin((2x)/2)})/(cos((2x)/2)-cos((4x)/2)`

`=(sin2x+sinx-3sin2x+3sinx)/(cosx-cos2x)`

`=(4sinx-2sin2x)/(cosx-(cos^2x-sin^2x))`

`=(4sinx-4sinx*cosx)/(cosx-cos^2x+sin^2x)`

`=(4sinx(1-cosx))/(cosx(1-cosx)+(1-cosx)(1+cosx))`

`=(4sinx(1-cosx))/((1-cosx)(cosx+1+cosx)`

`=(4sinx)/(1+2cosx)=RHS`

Answer 3

`LHS=cot(A/2)-3cot((3A)/2)`

`=cos(A/2)/sin(A/2)-cos((3A)/2)/sin((3A)/2)-2cot((3A)/2)`

`=(sin((3A)/2)*cos(A/2)-cos((3A)/2)*sin(A/2))/(sin(A/2)*sin((3A)/2))-2cot((3A)/2)`

#=sin(A)/(sin(A/2)*sin((3A)/2)) -2cot((3A)/2)#

#=(2sin(A/2)cos(A/2))/(sin(A/2)*sin((3A)/2)) -2cot((3A)/2)#

#=2cos(A/2)/sin((3A)/2)-2*cos((3A)/2)/sin((3 A)/2) #

#=2[(cos(A/2)-cos((3A)/2))/sin((3 A)/2)] #

#=2[(2sin(A/2)sin(A))/(3sin( A/2)-4sin^3(A/2))] #

#=(4sin(A/2)sin(A))/(sin(A/2)(3-4sin^2(A/2)) #
#=(4sin(A))/(3-2(1-cosA)) #

`=(4sin(A))/(1+2cosA)=RHS`