`tan theta + tan 2theta + tan 3theta = 0` को हल कीजिए |

Explanation

चूँकि `" "tan theta + tan 2theta + tan 3 theta = 0 " "` [दिया है]
`rArr " " tan theta + tan 2 theta + tan (theta + 2theta)=0`
`rArr " " tan theta + tan 2theta + (tan theta + tan 2theta)/(1-tan theta tan 2theta)=0`
`rArr " " (tan theta + tan 2theta)[1+(1)/(1-tan theta tan 2theta)]=0`
`rArr " " (tan theta + tan 2theta)[(1-tan theta tan 2theta + 1)/(1-tan theta tan 2theta)]=0`
`rArr " " (tan theta + tan 2theta)[(2-tan theta tan 2theta)/(1-tan theta tan 2theta)]=0`
यदि `" " tan theta + tan 2theta = 0`
`rArr " " tan 2theta =-tan theta = tan(-theta)`
`rArr " " 2theta = n pi - theta`
`rArr " " 3theta = n pi`
`rArr " " theta = (1)/(2)n pi " "` (जहाँ `n in I`)
और यदि `(2-tan theta tan 2theta)/(1-tan theta tan 2theta)=0`
`rArr " " 2 - tan theta tan 2theta = 0`
`rArr " " tan theta tan 2theta =2`
`rArr " " tan theta .(2tan theta)/(1-tan^(2)theta)=2`
`rArr " " (tan^(2)theta)/(1-tan^(2)theta)=1`
`rArr " " tan^(2)theta = 1-tan^(2)theta`
`rArr " " 2tan^(2)theta =1`
`rArr " " tan^(2)theta = (1)/(2) = tan^(2)alpha " "` [मान लीजिए]
`rArr " " tan theta = pm tan alpha = tan (pm alpha)`
`rArr " " theta = n pi pm alpha`
जहाँ `" " tan alpha = (1)/(sqrt(2))` एवं `n in I`
अतः, `" " theta` का अभीष्ट मान `=(n pi)/(3)`, जहाँ `n in I`
अथवा `" " n pi + alpha`, जहाँ `tan alpha = (1)/(sqrt(2))` एवं `n in I`.