Question

If $2-{\cos }^2\theta =3\sin \theta \cos \theta$ , $\sin \theta \ne cos\theta$ , then find the value of $\cot \theta$

• A. $\frac{1}{2}$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

Answer: (D)

$2-{\cos }^2\theta =3\sin \theta \cos \theta$
Dividing both sides by ${\cos }^2\theta$ ,
we get $2{\sec }^2\theta -1=3\tan \theta$
$2(1+{\tan }^2\theta )-1=3\tan \theta$
$\Rightarrow$ $2+2{\tan }^2\theta -1=3\tan \theta$
$\Rightarrow$ $2{\tan }^2\theta -3\tan \theta +1=0$
$\Rightarrow$ $2{\tan }^2\theta -2\tan \theta -\tan \theta +1=0$
$\Rightarrow$ $2\tan \theta (\tan \theta -1)-(\tan \theta -1)=0$
$\Rightarrow$ $(2\tan \theta -1)(\tan \theta -1)=0$
$\Rightarrow$ $\tan \theta =\frac{1}{2}$ and $1$
$\Rightarrow$ $\cot \theta =2$ and $1$