Question

State true or false.

If$\left(1+{\sin }^2\theta \right)=3\sin \theta \cos \theta$ then $\tan \theta =-1or\frac{1}{2}.$

• A. True
• B. False

Answer 1

The correct option is B False

$(1+{\sin }^2\theta )=3\sin \theta \cos \theta$

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

$\Rightarrow 2+1-\cos 2\theta =3\sin 2\theta$

$\Rightarrow 3-\cos 2\theta =3\sin 2\theta$

$\Rightarrow 3-\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }=3\times \frac{2\tan \theta }{1+{\tan }^2\theta }$

$\Rightarrow \frac{3+3{\tan }^2\theta +1-{\tan }^2\theta }{1+{\tan }^2\theta }=\frac{6\tan \theta }{1+{\tan }^2\theta }$

$\Rightarrow 3+3{\tan }^2\theta +1-{\tan }^2\theta =6\tan \theta$

$\Rightarrow 4+2{\tan }^2\theta -6\tan \theta =0$

$\Rightarrow 2{\tan }^2\theta -6\tan \theta +4=0$

$\Rightarrow {\tan }^2\theta -3\tan \theta +2=0$

$\Rightarrow (\tan \theta -1)(2\tan \theta -1)=0$

$\Rightarrow \tan \theta =1$ or $\frac{1}{2}$

Hence the given statement is false.