Question

If $2{\sin }^2\theta +{\cos }^2{45}^o=\tan {45}^o$ and $0\le \theta \le {90}^o$, then $\tan \theta$ equals

• A. $\sqrt{3}$
• B. $\frac{1}{\sqrt{3}}$
• C. $1$
• D. $\infty$

Answer 1

Answer: (B)

$\frac{1}{\sqrt{3}}$

$2{\sin }^2\theta +{\cos }^2{45}^o=\tan {45}^o\Rightarrow 2{\sin }^2\theta +\frac{1}{2}=1\Rightarrow 2{\sin }^2\theta =1-\frac{1}{2}\Rightarrow {\sin }^2\theta =\frac{1}{4}\Rightarrow \sin \theta =\frac{1}{2}\Rightarrow \theta ={30}^o$

Now, $\tan \theta =\tan {30}^o=\frac{1}{\sqrt{3}}$

Therefore, Answer is $\frac{1}{\sqrt{3}}$