Question

If $2{\sin }^2\theta +{\cos }^2{45}^0=\tan {45}^{0}and0\le \theta \le {90}^0,$ then the value of tan $\theta$ is

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

Answer 1

The correct option is C $\frac{1}{\sqrt{3}}$

$Given{0}^{\circ }\le \theta \le {90}^{\circ }.Soallthetrigonometricratiowillbepositive.Given2{sin}^2\theta +{cos}^2{45}^{\circ }=tan{45}^{\circ }or2{sin}^2\theta +\frac{1}{2}=1or{sin}^2\theta \frac{1}{4}\therefore {cos}^2\theta =1-{sin}^2\theta =1-\frac{1}{4}=\frac{3}{4}\therefore {tan}^2\theta =\frac{{sin}^2\theta }{{cos}^2\theta }=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}\therefore tan\theta =\frac{1}{\sqrt{3}}Ans$