Question

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

Answer 1

Solve for the required proof

Given: $1+{\sin }^2\left(\mathrm{\theta }\right)=3\sin \left(\mathrm{\theta }\right)\cos \left(\mathrm{\theta }\right)$

$\Rightarrow$ $\left[{\sin }^2\left(\mathrm{\theta }\right)+{\cos }^2\left(\mathrm{\theta }\right)\right]+{\sin }^2\left(\mathrm{\theta }\right)=3\sin \left(\mathrm{\theta }\right)\cos \left(\mathrm{\theta }\right)$ $\left[\because {\sin }^2\left(\mathrm{\theta }\right)+{\cos }^2\left(\mathrm{\theta }\right)=1\right]$

$\Rightarrow$ $2{\sin }^2\left(\mathrm{\theta }\right)+{\cos }^2\left(\mathrm{\theta }\right)=3\sin \left(\mathrm{\theta }\right)\cos \left(\mathrm{\theta }\right)$

$\Rightarrow$ $\frac{2{\sin }^2\left(\mathrm{\theta }\right)}{{\cos }^2\left(\mathrm{\theta }\right)}+\frac{{\cos }^2\left(\mathrm{\theta }\right)}{{\cos }^2\left(\mathrm{\theta }\right)}=\frac{3\sin \left(\mathrm{\theta }\right)\cos \left(\mathrm{\theta }\right)}{{\cos }^2\left(\mathrm{\theta }\right)}$ [Dividing both sides by ${\cos }^2\left(\mathrm{\theta }\right)$]

$\Rightarrow$ $2{\tan }^2\left(\mathrm{\theta }\right)+1=\frac{3\sin \left(\mathrm{\theta }\right)}{\cos \left(\mathrm{\theta }\right)}$ $\left[\because \frac{\sin \left(\mathrm{\theta }\right)}{\cos \left(\mathrm{\theta }\right)}=\tan \left(\mathrm{\theta }\right)\right]$

$\Rightarrow$ $2{\tan }^2\left(\mathrm{\theta }\right)+1=3\tan \left(\mathrm{\theta }\right)$

$\Rightarrow$ $2{\tan }^2\left(\mathrm{\theta }\right)-3\tan \left(\mathrm{\theta }\right)+1=0$

$\Rightarrow$ $2{\tan }^2\left(\mathrm{\theta }\right)-2\tan \left(\mathrm{\theta }\right)-\tan \left(\mathrm{\theta }\right)+1=0$

$\Rightarrow$$2\tan \left(\mathrm{\theta }\right)\left(\tan \left(\mathrm{\theta }\right)-1\right)-1\left(\tan \left(\mathrm{\theta }\right)-1\right)=0$

$\Rightarrow$ $\left(2\tan \left(\mathrm{\theta }\right)-1\right)\left(\tan \left(\mathrm{\theta }\right)-1\right)=0$

$\Rightarrow$ $\tan \left(\mathrm{\theta }\right)=1$ or $\frac{1}{2}$

Hence proved.