Question

If sin²θ = 3 sinθcosθ , prove that tanθ = 1 or ½.​

Answer 1

Step-by-step explanation:

Correct Question:-

If 1 + sin²θ = 3 sinθcosθ , prove that tanθ = 1 or ½.

Given:-

• sin²θ = 3 sinθcosθ

To Prove:-

• tanθ = 1 or ½

Proof:-

$\rm1 + {sin}^{2} \theta = 3sin \theta cos \theta$

Dividing LHS and RHS by cos²θ

$\rm \dashrightarrow\dfrac{1 + {sin}^{2} \theta }{ {cos}^{2} \theta} = \dfrac{3sin \theta cos \theta}{ {cos}^{2} \theta}$

$\rm \dashrightarrow\dfrac{1 } { {cos}^{2} \theta} + \dfrac{ {sin}^{2} \theta }{ {cos}^{2} \theta} = \dfrac{3sin \theta }{ {cos} \theta}$

$\rm \dashrightarrow {sec}^{2} \theta+ {tan}^{2} \theta = 3tan \theta \: \: \: \: \: \: \: \bigg( \because \dfrac{1}{ {cos}^{2} \theta } = {sec}^{2} \theta \: \: \: (and) \: \: \dfrac{sin \theta}{cos \theta} = tan \theta \bigg)$

$\rm \dashrightarrow1 + {tan}^{2} \theta+ {tan}^{2} \theta = 3tan \theta \: \: \: \: \: \: \bigg( \because {sec}^{2} \theta = 1 + { tan }^{2} \theta \bigg)$

$\rm \dashrightarrow 2 {tan}^{2}\theta - 3tan \theta + 1 = 0$

$\rm \underline{By \: factorisation:-}$

$\rm \dashrightarrow 2 {tan}^{2}\theta - 2tan \theta - tan \theta+ 1 = 0$

$\rm \dashrightarrow 2tan \theta( {tan}\theta - 1) - 1( tan \theta - 1) = 0$

$\rm \dashrightarrow (2tan \theta - 1)( tan \theta - 1) = 0$

$\rm \dashrightarrow 2tan \theta - 1 = 0 \: \: \: \: (or) \: \: \: \: tan \theta - 1= 0$

$\rm \dashrightarrow tan \theta = \dfrac{1}{2} \: \: \: \: (or) \: \: \: \: tan \theta = 1$

$\rm \dashrightarrow tan \theta = \dfrac{1}{2} \: \: (or) \: \: 1$

Hence Proved