pi/4 dude...............wait il post d solution........
let tanθ= limx-0(cosx+sinx/cosx+sinx.cosx)^{cosecx} then, θ=?
a)pi/2 b)pi/3 c)pi/6 d)pi/4
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6 Answers
Akand
·2009-03-17 07:53:42
tanθ=e(cosx+tanx+sinxcosx-1)cosecx
=e(tanx+sinx)/sinx
=e2..............wat d heck sorry made a msitake
wel ok.......
=e(tanx+sinx)/sinx
=e(tanx+sinx)cosecx
=e0
=1
so pi/4........
Optimus Prime
·2009-03-18 00:12:45
lim→0 (cosx+sinx/cosx+sixcosx)cosecx
lim→0 (1+ sinx-cosx/cosx+sinxcosx)cosecx
elim→0 (sinx-sinxcosx/cosx+sinxcosx)cosecx
e0= 1
tanA= 1
A= pi/4