eliminate θ from the foll eqns.:
cosecθ - sinθ = m
secθ - cosθ=n
i couldnt do this! :(
49pata hai ki its very easy,....
i had solved it earlier...
but can't do now at ne cost! :(
will u plz show the soln?? please??
1cosecθ - sinθ = m
or, cosec2θ + sin2θ - 2 = m2 (By squaring) ............(i)
secθ - cosθ = n
or, sec2θ + cos2θ - 2 = n2 (By squaring) .............................(ii)
(i)+(ii)
=> cosec2θ + sec2θ + sin2θ + cos2θ - 4 = m2+ n2
=> cosec2θ + sec2θ + 1 - 4 = m2+n2
=> cosec2θ + sec2θ - 3 = m2+n2
=> sin2θ + cos2θsin2θcos2θ - (m2+n2) = 3
=> 1sin2θcos2θ - (m2+n2) = 3..........................(iii)
Now,
cosecθ - sinθ = m
or, 1-sin2θsinθ = m
or, cos2θsinθ = m...........(a)
Similarly, sin2θcosθ = n......(b)
Multiplying (a) and (b) , we get sinθcosθ = mn
Substitute the values of sinθcosθ in (iii) and simplify to get the answer... [12]
Phew!
1for the first equation
√m2/3 + n2/3n2/3 - √n2/3m2/3+n2/3 = m
for the second equation
√m2/3 + n2/3m2/3 - √m2/3m2/3+n2/3 = n
subho are they correct...........
1@ Subho
@ Nihal
@ everyone else...
When we are given a problem like "eliminate θ from these two equations, does that mean we have to frame a single equation or two separate equations (in which θ is eliminated) ?
The way I did was by forming a joint degree-two equation...
Nihal formed two linear equations....
Which is the required / preferred method ?
49meaning is same!
u have to find relation b/w m and n here!
so u give 1 relation or 2its all the same!
like if u tell he is my grandfather or say he is my father's father its all the same!! [1]