The general solution of `tan theta+tan 2 theta+tan 3 theta=0` is

Options

A. `theta = npi//6, n in Z` only
B. `theta=n pi pm alpha, n in Z`, where `tan alpha=1//sqrt(2)` only
C. Both a and b
D. none of these

Answer

B

Explanation

From the given equation, we have
`tan theta + tan 2 theta + tan (theta + 2 theta)=0`
`rArr (tan theta + tan 2 theta)+ (tan theta + tan 2 theta)/(1-tan theta tan 2 theta)=0`
`rArr (tan theta + tan 2 theta) [1+1/(1-tan theta tan 2 theta)]=0`
`rArr (tan theta + tan 2 theta) (1- tan theta tan 2 theta)=0`
`rArr tan theta = tan (-2 theta)`
or `2-tan theta [(2 tan theta)//(1-tan^(2) theta)]=0`
`rArr theta = n pi -2 theta or 1-2 tan^(2) theta =0`
`rArr theta = n pi //3 or tan^(2) theta = 1//2=tan^(2) alpha` (say)
Therefore, `theta=n pi pm alpha`, where `tan alpha =1//sqrt(2)`.
`( :'" for " theta =n pi//6, tan 3 theta" is not defined")`