limit x tends to 0 then what is the value of 1-cosmx/1-cosnx=

limit x tends to 0   1-cosmx/1-cosnx= now rationalise nx))/(1-cos²nx)(1+cosnx)= ((1-cosmx)x1+cosnx/)x(1+cosnx(1-cosmx/1-cos then solve that you get zerothere fore limit x tends to 0 then what is the value of 1-cosmx/1-cosnx=0

(1-cosmx/1-cosnx)x(1+cosnx/1+cosnx)= ((1-cosmx)x(1+cosnx))/(1-cos²nx)1-cosmx/1-cosnx= now rationalise tends to 0   limit xthen solve that you get zerothere fore limit x tends to 0 then what is the value of 1-cosmx/1-cosnx=0 sorry but the options are a.mn  b.m/n   c.m^2/n²    d.m-n       

limit x tends to zero (1-cosmx)/(1-cosnx) by dividing the euation with mx and multiply with nx then you will get =limit x tends to zero (1-cosmx)/(1-cos nx)*(nx)/(mx) on re arranging we will get

              =((mx)(1-cosmx)/(mx)/(nx)(1-cos nx)/(nx)  but we know that 1-cos x=2sin^2 x/2

              =(nx/mx)(2(2sin^2 (mx)/2)/(mx)/2)/2(2sin^2 (nx)/2)/(nx)/2)

              =(n/m)(16/16)(n/m) limit x tends to zero ((sin (mx)/2)/(mx)/2))sin (mx)/2)/(mx)/2))/sin^2 (nx)/2)/(nx)/2)sin^2 (nx)/2)/(nx)/2)

              =(n/m)^2

 so the answer is (n/m)^2