1Q2)
theta = A
delta = B
i got tan A = √3,tanB=1/√3
or (-√3,-1/√3)
just simpligy leftmost equality to get sin2A=sin2B
and then convert it into tan wala formula..and then the realtion from rightmost equality..
Q1 - If cos4 (θ + α) , sin4(θ + α) are the roots of the equation x2 +b(2x+1) = 0 and cos2(θ+β),sin2(θ+β) are the roots of the equation x2+4x+2 = 0 then b is equal to ?
Q2 - (Sinθ / sin δ)2 = tan θ / tanδ = 3 then tanδ = and tan θ = ?
Q3 - If secθ + tanθ = 1 then one root of equation a(b-c)x2 +b(c-a)+c(a-b) =0 is a.Tanθ b.secθ c.cosθ d.sinθ
Please show an explanation too !
Thanks a lot !!
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8 Answers
1Q1 ) seems to be wrong
sum of roots is not 1 =cos2(θ+β)+sin2(θ+β)
Q3) clearly, 1 is the root of the equation
also secθ + tanθ = 1 and secθ - tanθ = 1,add => secQ = 1 hence option (b)
13For the first question the book says the answers are -1 and 2
and in question 3 how can you say that secθ - tanθ = 1 ?
and in question 3 the book says that even cosθ is the answer.
3Yes you are right for the second question.Can you explain it a bit more ?
1sec2A - tan2A=1 so apply (a-b)(a+b)=a2-b2
a=secA
b=secB
now i think u can understand
and yes cos(theta) shld also be the answer as secA=1 automatically implies that cosA is also 1..right??
3Yeah okay understood.
What about the other 2 questions ? Anyone trying them ?
1arey 1 is wrong ..surely ..i have given u the reason, see if u have typed the question correctly?
for second
sin2A/sin2B = sinA/cosA *cosB/sinB
cancel out commomn terms
and get sin2A=sin2B
now,this implies
2tanA/1+tan2A = 2tanB/1+tan2B
also tanA=3tanB
6tanB/1+9tan2B = same
simplify to get tan2B = 1/3