(solved)Question 19 SSC-CGL 2018 June 4 Shift 1 | If $\displaystyle 4 - 2\sin^2 \theta - 5\cos \theta = 0$, with $\displaystyle 0\degree \lt \theta \lt 90 \degree$, then the value of $\displaystyle \sin \theta + \tan \theta$ is?

Parveen, Feb 11, 2020

Question 19
SSC-CGL 2018 June 4 Shift 1

If $\displaystyle 4 - 2\sin^2 \theta - 5\cos \theta = 0$, with $\displaystyle 0\degree \lt \theta \lt 90 \degree$, then the value of $\displaystyle \sin \theta + \tan \theta$ is?

$\displaystyle \frac{3\sqrt 3}{2}$
$\displaystyle 3\sqrt 2$
$\displaystyle \frac{3\sqrt 2}{2}$
$\displaystyle 2\sqrt 3$

Solution 1

From the given options we can make a guess that $\displaystyle \theta$ could be one of the standard angles $\displaystyle 30 \degree$, $\displaystyle 45 \degree$, $\displaystyle 60 \degree$ such that $\displaystyle \cos \theta$ doesn't involve a surd ($\displaystyle \sqrt {}$). We can straightway try and verify that $\displaystyle \theta = 60 \degree$ indeed is the correct value. Hence $\displaystyle \sin 60\degree + \tan 60 \degree$ = $\displaystyle \frac{\sqrt 3}{2} + \sqrt 3$ = $\displaystyle \frac {3\sqrt 3}{2} \:\underline{Ans}$

Solution 2

Solve the given equation first.

$ \begin{aligned} 4 - 2\sin^2 \theta - 5\cos \theta &= 0\\ 4 - 2 (1 - \cos^2 \theta) - 5\cos \theta &= 0\\ 2 \cos^2 \theta - 5\cos \theta + 2 &= 0\\ (2 \cos \theta - 1)(\cos \theta - 2) &= 0\\ \end{aligned} $

From the above, we get

$ \begin{aligned} \cos \theta &= \frac12\\ \implies \theta &= 60 \degree\\ \implies &\sin \theta + \tan \theta \\ &= \sin 60\degree + \tan 60 \degree \\ &=\frac {3\sqrt 3}{2}\:\underline{Ans} \end{aligned} $