Question

if 1+sin²θ = 3sinθcosθ then prove that tanθ = 1 or ½
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Answer 1

Answer:

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Answer 2

$\bigstar \underline{\underline{\sf Given:-}}$

• 1+sin²θ = 3sinθcosθ

$\bigstar \underline{\underline{\sf To\ prove:-}}$

• tanθ = 1 or ½

$\bigstar \underline{\underline{\sf Proof:-}}$

We need to know :

$\\ \longmapsto \sf cosec^2\theta =\frac{1}{sin^2\theta}$

$\longmapsto \sf cosec^2\theta - cot^2\theta = 1$

$\longmapsto \sf cot\theta =\frac{cos\theta}{sin\theta}$

$\longmapsto \sf tan\theta = \frac{1}{cot\theta}$

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Given, 1+sin²θ = 3sinθcosθ

Dividing both LHS&RHSwith sin²θ :

$\\ :\implies \sf \frac{1+sin^2\theta}{sin^2 \theta} =\frac{3sin\theta \ cos\theta}{sin^2 \theta} \\\\:\implies \sf \frac{1}{sin^2 \theta} +\frac{sin^2\theta}{sin^2 \theta} =\frac{3cos\theta}{sin\theta} \\\\:\implies \sf \frac{1}{sin^2 \theta} +1=\frac{3cos\theta}{sin\theta}\\\\:\implies \sf cosec^2\theta +1=3cot\theta \\\\:\implies \sf \bold{cosec^2\theta = cot^2\theta + 1}\\\\:\implies \sf cot^2\theta +1+1=3cot\theta\\\\:\implies \sf cot^2\theta +2=3cot\theta$

$:\implies \sf cot^2\theta -3 cot\theta +2 = 0$

Splitting the middle term,

$\\ :\implies \sf cot^2\theta -cot\theta -2cot\theta +2 =0\\\\:\implies \sf cot\theta(cot\theta -1)-2(cot\theta -1)=0\\\\:\implies \sf (cot\theta-1)(cot\theta -2)=0\\\\:\implies \sf cot\theta - 1=0 \ \ \ (or)\ \ \ cot\theta -2=0\\\\:\implies \sf \underline{\bold{cot\theta =1,2}}$

And,

$\implies \tt \bold{\frac{1}{cot\theta} =tan\theta }$

$:\implies \sf tan\theta =\frac{1}{1} \ \ or \ \ \frac{1}{2} \\\\\implies \ \boxed{\boxed{\bold{\orange{tan\theta = 1,\frac{1}{2} }}}}$

Hence proved.

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