If `1+sin^(2)theta = 3sinthetacostheta`, then prove that `tantheta=1`

1.	If `1+sin^(2)theta = 3sinthetacostheta`, then prove that `tantheta=1` or `1/2`.
Answer» Given, `1+sin^(2)theta= 3sintheta.costheta` On dividing by `sin^(2)theta` on both sides, we get `1/(sin^(2)theta)+1=3.cottheta` `[therefore cottheta=(costheta)/(sintheta)]` `rArr "cosec"^(2)theta + 1=3.cottheta` `["cosec"theta = 1/(sintheta)]` `rArr 1+cot^(2)theta+1=3.cottheta` `["cosec"theta= 1/(sintheta)]` `rArr 1+ cot^(2)theta+1 = 3.cottheta` `[therefore "cosec"^(2)theta-cot^(2)theta=1]` `rArr cot^(2)theta-3cottheta+2=0` `rArr cot^(2)theta - 2cottheta-cottheta+2=0` [by splitting the middle term] `rArr cottheta(cottheta-2)-1(cottheta-2)=0` `rArr (cottheta-2)(cottheta-1)=0 rArr cottheta=1` or 2 `rArr tantheta=1 or 1/2` `[therefore tantheta=1/(cottheta)]` Hence Proved.

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