Question
If $1+\sin^2\theta=3\sin\theta\cos\theta,$ then prove that $\tan\theta=1,\text{ or }\frac{1}{2}.$
✓
Answer
To solve an equation in we have to change it into one trigonometric ratio.
Given trigonometric equation is.
$1+\sin^2\theta=3\sin\theta\cos\theta$
$\Rightarrow\ \frac{1}{\sin^2\theta}+\frac{\sin^2\theta}{\sin^2\theta}=\frac{3\sin\theta\cos\theta}{\sin^2\theta}$ $[\text{Dividing by}\sin^2\theta\text{ both sides}]$
$\Rightarrow\ \text{cosec}^2\theta+1=3\cot\theta$
$\Rightarrow\ 1+\cot^2\theta+1-3\cot\theta=0$
$\Rightarrow\ \cot^2\theta-3\cot\theta+2=0$
$\Rightarrow\ \cot^2\theta-2\cot\theta-1\cot\theta+2=0$
$\Rightarrow\ \cot\theta(\cot\theta-2)-1(\cot\theta-2)=0$
$\Rightarrow\ (\cot\theta-2)(\cot\theta-1)=0$
$\Rightarrow\ \cot\theta-2=0\text{ or }(\cot\theta-1)=0$
$\Rightarrow\ \cot\theta=2\text{ or }\cot\theta=1$
$\Rightarrow\ \tan\theta=\frac{1}{2}\text{ or }\tan\theta=1$
Hence, $\tan\theta=\frac{1}{2},\text{ or }1.$
Given trigonometric equation is.
$1+\sin^2\theta=3\sin\theta\cos\theta$
$\Rightarrow\ \frac{1}{\sin^2\theta}+\frac{\sin^2\theta}{\sin^2\theta}=\frac{3\sin\theta\cos\theta}{\sin^2\theta}$ $[\text{Dividing by}\sin^2\theta\text{ both sides}]$
$\Rightarrow\ \text{cosec}^2\theta+1=3\cot\theta$
$\Rightarrow\ 1+\cot^2\theta+1-3\cot\theta=0$
$\Rightarrow\ \cot^2\theta-3\cot\theta+2=0$
$\Rightarrow\ \cot^2\theta-2\cot\theta-1\cot\theta+2=0$
$\Rightarrow\ \cot\theta(\cot\theta-2)-1(\cot\theta-2)=0$
$\Rightarrow\ (\cot\theta-2)(\cot\theta-1)=0$
$\Rightarrow\ \cot\theta-2=0\text{ or }(\cot\theta-1)=0$
$\Rightarrow\ \cot\theta=2\text{ or }\cot\theta=1$
$\Rightarrow\ \tan\theta=\frac{1}{2}\text{ or }\tan\theta=1$
Hence, $\tan\theta=\frac{1}{2},\text{ or }1.$
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