Question

If $0^{\circ }\le \theta \le {90}^{\circ }$ and $\sqrt{2}tan\theta -sec\theta =0$, then the value of $(\sqrt{2}sin\theta +2tan\theta )$ is -

• A. $\sqrt{2}+2$
• B. $\frac{\sqrt{2}}{2}+\frac{2}{\sqrt{3}}$
• C. 3
• D. 2

Answer 1

Answer: (C)

3

$\sqrt{2}\tan \theta =\sec \theta or2{\tan }^2\theta ={\tan }^2\theta +1$
$\therefore {\tan }^2\theta =1\therefore \theta ={45}^{\circ }$
$\sqrt{2}\sin \theta +2\tan \theta =\sqrt{2}\times \frac{1}{\sqrt{2}}+2\times 1=3$