Question

If $\sec \theta +\tan \theta =2,$ then find value of $\sin \theta$ where $0^{\circ }<0<{90}^{\circ }$

• A. $\frac{2}{5}$
• B. $\frac{3}{5}$
• C. $\frac{4}{5}$
• D. $1$

Answer 1

The correct option is B $\frac{3}{5}$

$\sec \theta +\tan \theta =2⟶\left(1\right)$

Now, ${\sec }^2\theta -{\tan }^2\theta =1$

$\Rightarrow (\sec \theta -\tan \theta )(\sec \theta +\tan \theta )=1$

$\Rightarrow 2(\sec \theta -\tan \theta )=1$

$\Rightarrow \sec \theta -\tan \theta =\frac{1}{2}⟶\left(2\right)$

Adding equations $\left(1\right)\&\left(2\right),$ we get

$\Rightarrow 2\sec \theta =2+\frac{1}{2}=\frac{5}{2}$

$\Rightarrow \sec \theta =\frac{5}{4}$

Or $\cos \theta =\frac{4}{5}$

Now, $\sin \theta =\sqrt{1-{\cos }^2\theta }$

$=\sqrt{1-{\left(\frac{4}{5}\right)}^2}$

$=\sqrt{1-\frac{16}{25}}$

$=\sqrt{\frac{9}{25}}$

$=\frac{3}{5}$

Hence, the answer is $\frac{3}{5}.$