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Question

# If tanx+2tan2x+4tan4x+8cot8x=√3 then the general solution of x=

A
nπ+π3,Z
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B
nπ+π6,Z
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C
nπ+π4,Z
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D
nπ,Z
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Solution

## The correct option is B nπ+π6,∀∈Z tanx+2tan2x+4tan4x+8cot8x=√3 formulaused tanx−cotx =tanx−1tanx =tan2x−1tanx =−2[1−tan2x2tanx] =−2[12tanx1−tan2x] =−2[1tan2x] =−2cotx tanx=tanα x=nπ+α tanx+2tan2x+4tan4x+8cot8x=√3 addingandsubtractingcotx tanx−cotx+2tan2x+4tan4x+8cot8x+cotx=√3 ⇒−2cot2x+2tan2x+4tan4x+8cot8x+cotx=√3 ⇒2[tan2x−cot2x]+4tan4x+8cot8x+cotx=√3 ⇒2[−2cot4x]+4tan4x+8cot8x+cotx=√3 ⇒−4cot4x+4tan4x+8cot8x+cotx=√3 ⇒4[tan4x−cot4x]+8cot8x+cotx=√3 ⇒4[−2cot8x]+8cot8x+cotx=√3 ⇒−8cotx+8cotx+cotx=√3 ⇒cotx=√3 ⇒1tanx=√3 ⇒tanx=1√3=tan(π6) x=nπ+π6

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