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1.

Prove that: `cos18^0-si n 18^0=sqrt(2)sin27^0`

Answer» `L.H.S. = cos 18^@ - sin 18^@`
`= cos 18^@ - sin (90-72)^@`
`= cos 18^@ - cos72^@`
Using `cosA - cosB = -2sin((A-B)/2)((A+B)/2),`
`= -2sin((18-72)/2)^@sin((18+72)/2)^@`
`= -2sin(-27^@)sin45^@`
`=2sin27^@(1/sqrt2)`
`=sqrt2sin27^@ = R.H.S.`
Discussion
2.

Prove that: `(sin(A-C)+2sinA+sin(A+C)) / ("sin"("B"-"C")+2sinB+"sin"(B+C))`=`(sinA)/(sinB)``

Answer» `L.H.S. = (sin(A-C)+2sinA+sin(A+C)) / (sin(B-C)+2sinB+sin(B+C))`
`=(sinAcosC-cosAsinC+2sinA+sinAcosC+cosAsinC)/(sinBcosC-cosBsinC+2sinB+sinBcosC+cosBsinC)`
`=(2sinA(1+cosC))/(2sinB(1+cosC))`
`=sinA/sinB = R.H.S.`
Discussion
3.

Prove that: `(sinA+s in3A+sin5A+sin7A)/(cosA+cos3A+cos5A+cos7A)=tan4A`

Answer» `L.H.S = (sinA+sin3A+sin5A+sin7A)/(cosA+cos3A+cos5A+cos7A)`
`=((sin7A+sinA)+(sin5A+sin3A))/((cos7A+cosA)+(cos5A+cos3A)`
`=((2sin4Acos3A)+(2sin4AcosA))/((2cos4Acos3A+2cos4AcosA))`
`=(2sin4A)/(2cos4A)[(cos3A+cosA)/(cos3A+cosA)]`
`=tan4A = R.H.S.`
Discussion
4.

Prove that: `1+cos2x+cos4x+cos6x=4cosxcos2xcos3x`

Answer» `L.H.S. = 1+cos2x+cos4x+cos6x`
`=(1+cos4x)+(cos2x+cos6x)`
`=(2cos^2 2x)+(2cos((2x+6x)/2)cos((2x-6x)/2))`
`=(2cos^2 2x)+(2cos(4x)cos(2x))`
`=2cos 2x[cos 2x+cos 4x]`
`=2cos 2x[2cos3xcosx]`
`=4cosxcos2xcos3x = R.H.S.`
Discussion
5.

If `(sin(theta+alpha))/(cos(theta-alpha))=(1-m)/(1+m)`, prove that `tan(pi/4-theta)tan(pi/4-alpha)=m`

Answer» Let, `pi/4 - theta = A and pi/4-alpha = B`
Then`pi/4-theta+pi/4-alpha = A+B`
`=>theta+alpha = pi/2-(A+B)`
`=>sin(theta+alpha) = sin(pi/2-(A+B))`
`=>sin(theta+alpha) = cos(A+B)`
Similarly, `cos(theta-alpha) = cos(B-A)`
Now, putting these values in the given equation,
`(cos(A+B))/(cos(B-A)) = (1-m)/(1+m)`
`=>(cos(B-A))/(cos(A+B)) = (1+m)/(1-m)`
Using componendo and dividendo,
`=>(cos(B-A)+cos(A+B))/(cos(B-A)-cos(A+B)) = (1+m+1-m)/(1+m-1+m)`
`=>(2cosAcosB)/(2sinAsinB) = 2/(2m)`
`=>1/(tanAtanB) = 1/m`
`=>tanAtanb = m`
`:. tan(pi/4-theta)tan(pi/4-alpha) = m.`
Discussion
6.

Prove that: `4"cos"12^0cos48^0cos72^0=cos36^0`

Answer» `L.H.S. = 4cos12^@cos48^@cos72^@`
`=2*(2cos12^@cos48^@)cos72^@`
We know, `2cosAcosB = cos(A+B)+cos(A-B)`
So, our expression becomes,
`=2*(cos60^@+cos36^@)cos72^@`
`=2cos60^@cos72^@+2cos36^@cos72^@`
`=2(1/2)cos72^@+cos108^@+cos36^@`
`=cos72^@+cos(pi-72)^@+cos36^@`
`=cos72^@-cos72^@+cos36^@`
`=cos36^@`
`=R.H.S.`
Discussion
7.

If `sinx+siny=sqrt(3)(cos y-cos x),`then `sin3x+sin3y=`(a)`2sin3x`(b) 0 (c) 1 (d) none of these

Answer» `sinx+siny = sqrt3(cosy-cosx)`
`=>2sin((x+y)/2)cos((x-y)/2)= sqrt3(2sin((x+y)/2)sin((x-y)/2)) `
`=>sin((x+y)/2)cos((x-y)/2) - sqrt3sin((x+y)/2)sin((x-y)/2) = 0`
`=>sin((x+y)/2)[cos((x-y)/2) - sqrt3sin((x-y)/2)] = 0`
`=>sin((x+y)/2) = 0 or cos((x-y)/2) - sqrt3sin((x-y)/2) = 0`
`=>x+y = 0 or cot ((x-y)/2) = sqrt3`
`=>x = -y or (x-y)/2 = pi/6`
`=>x= -y or x = y+pi/3`
When `x = -y,`
`sin3x+siny = sin(-3y)+sin3y = sin3y-sin3y= 0`
When, `x = y+pi/3,`
`sin3x+siny = sin3(y+pi/3)+sin3y = sin(pi+3y)+sin3y= -sin3y+sin3y = 0`
`:.` In both cases, ` sin3x+sin3y = 0.`
Discussion
8.

If `cos(alpha+beta)*sin(gamma+delta)=cos(alpha-beta)*sin(gamma-delta),`prove that `cotalphacotbetacotgamma=cotdelta`

Answer» `cos(alpha+beta)sin(gamma+delta) = cos(alpha-beta)sin(gamma-delta)`
`cos(alpha+beta)/cos(alpha-beta) = sin(gamma-delta)/sin(gamma+delta)`
Using componendo and dividendo,
`=>(cos(alpha+beta)+cos(alpha-beta))/ (cos(alpha+beta)-cos(alpha-beta))= (sin(gamma-delta)+sin(gamma+delta))/(sin(gamma-delta)-sin(gamma+delta))`
`=>(2cosalphacosbeta)/(-2sinalphasinbeta) = (2singammacos(-delta))/(2sin(-delta)cosgamma)`
`=>(cosalphacosbeta)/(-sinalphasinbeta) = (singammacosdelta)/(-sindeltacosgamma)`
`=>cotalphacotbeta = cotdelta/cotgamma`
`=>cotalphacotbeta cotgamma = cot delta`
Discussion
9.

If `xcostheta=ycos(theta+(2pi)/3)=z cos(theta+(4pi)/3)`, prove that `x y+y z+z x=0.`

Answer» Let `xcostheta = ycos(theta+(2pi)/3) = zcos(theta+(4pi)/3) = k`
Then,
`k/x = cos theta, k/y =cos(theta+(2pi)/3), k/z = cos(theta+(4pi)/3)`
`:.k(1/x+1/y+1/z) = cos theta +cos(theta+(2pi)/3) + cos(theta+(4pi)/3)`
`=>k(1/x+1/y+1/z) = (cos theta +cos(theta+(4pi)/3)) + cos(theta+(2pi)/3)`
`=>k(1/x+1/y+1/z) = (2cos(theta+(2pi)/3)cos((-2pi)/3)) + cos(theta+(2pi)/3)`
`=>k(1/x+1/y+1/z) = (2cos(theta+(2pi)/3)cos((2pi)/3)) + cos(theta+(2pi)/3)`
`=>k(1/x+1/y+1/z) = (2cos(theta+(2pi)/3)(-1/2)) + cos(theta+(2pi)/3)`
`=>k(1/x+1/y+1/z) = -cos(theta+(2pi)/3) + cos(theta+(2pi)/3)`
`=>1/x+1/y+1/z = 0`
`=>(yz+zx+xy)/(xyz) = 0`
`=>xy+yz+zx = 0`
Discussion