IIT JEE , AIEEE Forum - Mathematics , Trignometry

Arjun

heres da solution for ue first question............

tanx=ntany......i hav taken x and y for simplicity.........

cos(x-y)=cosxcosy+sinxsiny

cos(x-y)=cosxcosy(1+tanxtany) as sinx/cosx= tanx

cos(x-y)=cosxcosy(1+ntan²y)

sec²(x-y)=sec²xsec²y/(1+ntan²y)²

=(1+tan²y)(1+tan²x)/(1+ntan²y)²

= (1+n²tan²y)(1+tan²y)/(1+ntan²y)²

1+(n-1)²tan²y/(1+ntan²y)²

(1+ntan²y/2)>ntan²y

tan²y/(1+ntan²y)²<1/4n

so substituting in da above obtained xpression..........

sec²(x-y)-1<(n-1)²/4n

tan²(x-y)<(n-1)²/4n

so the answer is

(n-1)²/4n

so all dis problem asks is just a play of formulae..........

enjoyed solvin it............

thanx for givin a nice problem gal........