You should use the following formula to prove that expression `tan(pi+x) = tan x` holds, such that:
`tan(a+b) = (tan a+tan b)/(1 - tan a*tan b)`
Reasoning by analogy yields:
`tan(pi+x) = (tan pi+tan x)/(1 - tan pi*tan x)`
Substituting 0 for `tan pi` yields:
`tan(pi+x) = (0+tan x)/(1 -...
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You should use the following formula to prove that expression `tan(pi+x) = tan x` holds, such that:
`tan(a+b) = (tan a+tan b)/(1 - tan a*tan b)`
Reasoning by analogy yields:
`tan(pi+x) = (tan pi+tan x)/(1 - tan pi*tan x)`
Substituting 0 for `tan pi` yields:
`tan(pi+x) = (0+tan x)/(1 - 0*tan x)`
`tan(pi+x) = (tan x)/1 => tan(pi+x) = tan x`
Hence, using the formula `tan(a+b) = (tan a+tan b)/(1 - tan a*tan b)` to evaluate the given expression yields that `tan(pi+x) = tan x` is valid.