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math doubts [NEW]

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astarmathsandphysics:
1)b^2-4ac=k^2-4*1*(-k+3)>0 and -k>sqrt(k^2-4*1*(-k+3))>0
k^2+4k-12=(k+6)(k-2)>0 so k>6 or k<-2 and k^2>k^2-4*1*(-k+3) so-k+3>0 so k<3

so k>6 or k<-2 anf k<3ie k<-2

")b^2-4ac<0 since y=0 has no roots
(3q)^2-4pr<0 so 4q^2<4pr so q^2<pr.

zara:
NEW QUESTIONS! PLZ HELP!


Q1)Given that x=sin-1(2/5), find the exact value of
(i) cos2x
(ii)tan2x

zara:
Q2) Prove this identity

1/cos(theta) + tan(theta) = cos(theta)/1-sin(theta)

Ghost Of Highbury:
first find x

sin-1 2/5 = 23.5

cos2 x = 0.84

and

tan2x = 0.19

just substitute and find it
---
next

(1/cos x + sinx/cosx) = (1+sinx/ cos x) * (1-sinx/1-sinx) = 1-sin2x/cosx - sinxcosx = cos2x/cosx-sinxcosx

= cos2x / cosx(1-sinx) = cosx/1-sinx

x = theta

LHS = RHS

hence proved

zara:
Q3)Given that the equation


[sin(theta) - cos(theta)]/[sin(theta)+cos(theta)] =6sin(theta)/cos(theta),

find the exact values of tan(theta).Hence find (theta) for 0<=(theta)<=360

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