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1.

Prove that:`sin^2A=cos^2(A-B)+cos^2B-2cos(A-B)cosAcosBdot`

Answer» `R.H.S. = cos^2(A-B)+cos^2B-2cos(A-B)cosAcosB`
`=cos^2B+cos(A-B)[cos(A-B)-2cosAcosB]`
`=cos^2B+cos(A-B)[cosAcosB+sinAsinB-2cosAcosB]`
`=cos^2B+cos(A-B)[-cosAcosB+sinAsinB]`
`=cos^2B-cos(A-B)[cosAcosB-sinAsinB]`
`=cos^2B-cos(A-B)[cos(A+B)]`
Now, we know, `cos(A-B)cos(A+B) = cos^2A-sin^2B`
So, it becomes,
`=cos^2B-cos^2A+sin^2B`
`=cos^2B+sin^2B-cos^2A`
`=1-cos^2A`
`=sin^2A = L.H.S.`
Discussion
2.

Prove that: `sin^2(pi/8+A/2)-sin^2(pi/8-A/2)=1/sqrt(2)sinA`

Answer» `L.H.S. = sin^2(pi/8+A/2) - sin^2(pi/8-A/2)`
`=(sin(pi/8)cos(A/2)+cos(pi/8)sin(A/2))^2-(sin(pi/8)cos(A/2)-cos(pi/8)sin(A/2))^2`
As `a^2-b^2 = (a+b)(a-b)`, so it becomes
`=(sin(pi/8)cos(A/2)+cos(pi/8)sin(A/2)+sin(pi/8)cos(A/2)-cos(pi/8)sin(A/2))(sin(pi/8)cos(A/2)-cos(pi/8)sin(A/2)-sin(pi/8)cos(A/2)+cos(pi/8)sin(A/2))`
`=4(sin(pi/8)cos(A/2)(cos(pi/8)sin(A/2))`
`=(2sin(pi/8)cos(pi/8))(2sin(A/2)cos(A/2))`
`=sin(2*pi/8)sin(2*A/2)`
`=sinpi/4sinA`
`=1/sqrt2sinA = R.H.S.`
Discussion
3.

If `2tanbeta+cotbeta=tanalpha`, prove that `cotbeta=2tan(alpha-beta)dot`

Answer» `R.H.S. = 2tan(alpha-beta) = (2(tanalpha - tanbeta))/(1+tanalphatanbeta)`
`=(2(2tanbeta+cotbeta- tanbeta))/(1+(2tanbeta+cotbeta)tanbeta)`
`=(2(tanbeta+cotbeta))/(1+(2tan^2beta+cotbetatanbeta)`
`=(2(tanbeta+1/tanbeta))/(1+(2tan^2beta+1/tanbetatanbeta)`
`=(2((tan^2beta+1)/tanbeta))/(2(tan^2beta+1))`
`=1/tanbeta = cotbeta = L.H.S.`
Discussion
4.

If `cos(alpha-beta)+cos(beta-gamma)+cos(gamma-alpha)=-3/2`, prove that`cosalpha+cosbeta+cosgamma=sinalpha+sinbeta+singamma=0`

Answer» `cos(alpha-beta)+cos(beta-gamma)+cos(gamma-alpha) = -3/2`
`=>2(cosalphacosbeta+sinalphasinbeta)+2(cosbetacosgamma+sinbetasingamma)+2(cosalphacosgamma+sinalphasingamma) = -3`
`=>2(cosalphacosbeta+cosbetacosgamma+cosalphacosgamma)+2(sinalphasinbeta+sinbetasingamma+sinalphasingamma)+3 = 0`
`=>2(cosalphacosbeta+cosbetacosgamma+cosalphacosgamma)+2(sinalphasinbeta+sinbetasingamma+sinalphasingamma)+(cos^2alpha+sin^2alpha)+(cos^2beta+sin^2beta)+(cos^2gamma+sin^2gamma) = 0`
`=>(cos^2alpha+cos^2beta+cos^2gamma +2(cosalphacosbeta+cosbetacosgamma+cosalphacosgamma))+(sin^2alpha+sin^2beta+sin^2gamma +2(sinalphasinbeta+sinbetasingamma+sinalphasingamma))=0`
`=>(cosalpha+cosbeta+cosgamma)^2+(sinalpha+sinbeta+singamma)^2 = 0`
`a^2+b^2 = 0` is possible onle when `a = b = 0.`
`:. cosalpha+cosbeta+cosgamma = sinalpha+sinbeta+singamma= 0`
Discussion
5.

If `tan(alpha+theta)=ntan(alpha-theta)`, show that `(n+1)sin2theta=(n-1)sin2alphadot`

Answer» `tan(alpha+theta) = n tan(alpha-theta)`
`=>tan(alpha+theta)/(tan(alpha-theta)) = n/1`
Using componendo and dividendo,
`=>(tan(alpha+theta)+tan(alpha-theta))/(tan(alpha+theta)-tan(alpha-theta)) = (n+1)/(n-1)`
`=>(sin(alpha+theta)/cos(alpha+theta)+sin(alpha-theta)/cos(alpha-theta))/(sin(alpha+theta)/cos(alpha+theta)-sin(alpha-theta)/cos(alpha-theta)) = (n+1)/(n-1)`
`=>(sin(alpha+theta)cos(alpha-theta)+cos(alpha+theta)sin(alpha-theta))/(sin(alpha+theta)cos(alpha-theta)+cos(alpha+theta)sin(alpha-theta)) =(n+1)/(n-1)`
`=>(sin(alpha+theta+alpha-theta))/(sin(alpha+theta-alpha+theta))=(n+1)/(n-1)`
`=>(sin2alpha)/(sin2theta) = (n+1)/(n-1)`
`=>(n-1)sin2alpha = (n+1)sin2theta.`
Discussion