Question

If 2 ${\sin }^2\Theta +{\cos }^2{45}^{\circ }=\tan \dot{{45}^{\circ }}and0\le \Theta \le {90}^{\circ }$ , then value of $\tan \Theta$ is

• A. $\sqrt{3}$
• B. 1
• C. $\frac{1}{\sqrt{3}}$
• D. not defined

Answer 1

Answer: (C)

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

$2si{n}^2\theta +co{s}^2{45}^{\circ }=tan{45}^{\circ }$

$tan{45}^{\circ }=1$ and $cos{45}^{\circ }=\frac{1}{\sqrt{2}}$

$2si{n}^2\theta +\frac{1}{2}=1$

$2si{n}^2\theta =\frac{1}{2}$

$si{n}^2\theta =\frac{1}{4}$

Since $0<\theta <90$

$sin\theta =\frac{1}{2}$

$\theta ={30}^{\circ }$

$\therefore tan{30}^{\circ }=\frac{1}{\sqrt{3}}$