If `"sin"(alpha + beta) "sin" (alpha-beta) = "sin" gamma(2"sin" beta + "sin"gamma) " where " 0 lt alpha, beta, lt pi,` then the straight line whose eq

L.H.S.=`|{:(alpha,alpha^(2),beta+gamma),(beta,beta^(2),gamma+alpha),(gamma,gamma^(2),alpha+beta):}|=|{:(alpha,alpha^(2),alpha+beta+gamma),(beta,beta^(2),alpha+beta+gamma),(gamma,gamma^(2),alpha+beta+gamma):}|` `(C_(3)toC_(3)+C_(1))` `=(alpha+beta+gamma)|{:(alpha,alpha^(2),1),(beta,beta^(2),1),(gamma,gamma^(2),1):}|` `(alpha+beta+gamma)|{:(alpha,alpha^(2),1),(beta-alpha,beta^(2)-alpha^(2),0),(gamma-alpha,gamma^(2)-alpha^(2),0):}|` `(R_(2)toR_(2)-R_(1),R_(3)toR_(3)-R_(1))` `(alpha+beta+gamma)(beta-alpha)(gamma-alpha)|{:(alpha,alpha^(2),1),(1,beta-alpha,0),(1,gamma-alpha,0):}|` `=-(alpha-beta)(gamma-alpha)(alpha+beta+gamma)1.|{:(1,beta+alpha),(1,gamma+alpha)` `=-(alpha-beta)(gamma-alpha)(alpha+beta+gamma)(gamma+alpha-beta-alpha)` `=(alpha-beta)(beta-gamma)(gamma-alpha)(alpha+beta+gamma)` =R.H.S.

Correct Answer - A Arrange the data as follows : `alpha-(7)/(2), alpha-3, alpha-(5)/(2),alpha-2, alpha-(1)/(2),alpha+(1)/(2),alpha+4,alpha+5` Median `=(1)/(2)` [value of 4th item+value of 5th item] `therefore " Median"=(alpha-2+alpha-(1)/(2))/(2)=(2alpha-(5)/(2))/(2)=alpha-(5)/(4)`

Correct Answer - C `cos^(2)alpha+cos^(2)beta+1-sin_(gamma)^(2)=1` `1-sin^(2)alpha+sin^(2)beta+1-sin^(2)gamma=1` `3-(sin^(2)alpha+sin^(2)beta+sin^(2)gamma)=1` `sin^(2)alpha+sin^(2)beta+sin^(2)gamma=3-1=2`

Correct Answer - A We have, `A(alpha, beta)^(-1)=1/e^(beta) [(e^(beta) cos alpha,-e^(beta) sin alpha,0),(e^(beta) sin alpha,e^(beta) cos alpha,0),(0,0,1)]` `=A(-alpha, -beta)`

Correct Answer - A::B::D `A=[(cos alpha,sin alpha,0),(cos beta,sin beta,0),(cos gamma,sin gamma,0)][(cos alpha,cos beta,cos gamma),(sin alpha,sin beta,sin gamma),(0,0,0)]` Clearly, A is symmetric and `|A|=0`, hence, singular and not invertiable. Also, `A A^(T)...

Correct Answer - C `(alpha^(3))/(2) cosec^(2) ((1)/(2) tan^(-1). (alpha)/(beta)) + (beta^(3))/(2) sec^(2) ((1)/(2) tan^(-1).(beta)/(alpha))` `= alpha^(3) (1)/(1 - cos(tan^(-1) ((alpha)/(beta)))) + beta^(3) (1)/(1 + cos (tan^(-1).(beta)/(alpha)))` `= alpha^(3) (1)/(1 -cos (cos^(-1)...

Correct Answer - C For `alpha=-pi//2,beta=pi//2andgamma=2pi` `sinalpha+sinbeta=singamma=-2` Hence, the minimum value of the expression is negative.

Correct Answer - B `(b)` `P(x)=x^(6)-x^(5)-x^(3)-x^(2)-x` and `Q(x)=x^(4)-x^(3)-x^(2)-1` Here` P(x)=Q(x)(x^(2)+1)+x^(2)-x+1` `:. sumP(alpha)=sumalpha^(2)-sumalpha+4` `=(sumalpha)^(2)-2sum(alphabeta)-sumalpha+4` `=1-2(-1)-(1)+4` `=6`

Correct Answer - A `(a)` `a=e^(ialpha)`, `b=e^(ibeta)`, `c=e^(igamma)` Clearly `a+b+c=0` `impliesa+b+c=0=(1)/(a)+(1)/(b)+(1)/(c )` `impliesa^(3)+b^(3)+x^(3)=3abc` `impliessin3alpha+sin3beta+sin3gamma=sin(alpha+beta+gamma)`

Correct Answer - D `(d)` We have `betaalpha+gammaalpha+alphabeta=0` `Delta=(1)/(alpha^(3)beta^(3)gamma^(3))|{:(betagamma,gammaalpha,alphabeta),(gammaalpha,alphabeta,betagamma),(alphabeta,betagamma,gammaalpha):}|` `=(1)/(alpha^(3)beta^(3)gamma^(3))|{:(betagamma+gammaalpha+alphabeta,gammaalpha,alphabeta),(gammaalpha+alphabeta+betagamma,alphabeta,betagamma),(alphabeta+betagamma+gammaalpha,betagamma,gammaalpha):}|` [using `C_(1)toC_(1)+C_(2)+C_(3)`] `=(1)/(alpha^(3)beta^(3)gamma^(3))|{:(0,gammaalpha,alphabeta),(0,alphabeta,betagamma),(0,betagamma,gammaalpha):}|=0` [all zero property].