The value of `2tan^(-1)(cos e ctan^(-1)x-tancot^(-1)x)` is equal to (a)`cot^(-1)x` (b) `cot^(-1)1/x` (c)`tan^(-1)x` (d) none of these

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tan(cot^(-1)x) is equal to

cot(tan^(-1)x+cot^(-1)x)

cos[tan^(-1){sin(cot^(-1)x)}] is equal to

sin[cot^(-1) {tan(cos^(-1)x)}] is equal to

tan^(-1)(cot x)-tan^(-1)(cot2x)=

tan^(-1)(cot x)+cot^(-1)(tan x)

cot^(-1)x=tan^(-1)x then

Prove that 2tan^(-1)(cos e ctan^(-1)x-tancot^(-1)x)=tan^(-1)x(x!=0)dot