urgent

LimÂ ( n â†’ infinity)Â {Â (1/2)Â tan x/2Â Â +Â (1/2²) tan x/2²Â + (1/2³) tan x/2³ .............+(1/2ⁿ)tan x/2ⁿ) }

6 Answers

use

tanx = cotx - 2cot2x

ull find succesive terms getting cancelled
ull end up with

cotx/2ⁿ/2ⁿ - cotx = 1 - cotx

don't you think it should be tanx=cotx-2cot2x

YES U R RITE SHRIYA..........

tanx = cotx - 2cot2x
ull find succesive terms getting cancelled

so at last u get
cotx/2ⁿ/2ⁿ -cot2x
1-cot2x.......

i think its rite....may hav done sum stupid mistakes

yes shriya i know that

just a typo

ive edited the ans now its right :)

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