1.prove that sinx.siny.sin(x-y) + siny.sinz.sin(y-z) + sinz.sinx.sin(z-x) + sin(x-y).sin(y-z).sin(z-x) = 0
2.f(α,β)= cos4α/cos2β + sin4α/sin2β then prove that f(α,β) = 1
3.cosA = tanB, cosB = tanC and cosC= tanA
then show that sinA = sinB = sinC = 2sin18
31.
LHS=
1/2siny.[2sinx.sin(x-y) +2 .sinz.sin(y-z)] + 1/2sin(z-x).[sinx.sinz + sin(x-y).sin(y-z)]
=1/2siny.[cosy-cos(2x-y)+cos(2z-y)-cosy]+1/2sin(z-x).[cos(x+z-2y)-cos(x-z)+cos(z-x)-cos(x+z)]
=1/2siny.2sin(x+z-y)sin(x-z)+1/2sin(z-x)sin(x+z-2y)siny
=0
11U have
sinB=cosAtanC ....(i)
cosB=tanC..........(ii)
Squaring and adding, u get cos2C=(1+cos2A)(1-cos2C).
Change the entire thing into sine to get 2sin4A-6sin2A+2=0......Now solve this to get the ans.
P.S - there should be 1 more constraint on the values. As far as I remember, this qsn is from fiitjee aits, where it was given A,B,C are the angles of a triangle.
1i think so 2nd problem is tooooo much easy
cos4[a]/cos2[.b]+sin4[a]/sin2[.b]
=> {cos4[a]sin2[.b]+sin4[a]sin2[.b]}/{cos2[.b]sin2[.b]}
take the lcm
and then take the factor {cos2[.b]sin2[.b]} common in numerator
you may get the answer
