AP ICET 2016 Paper

I Given : $\sec \mathit{\theta }+\mathrm{cosec}\mathit{\theta }=5$
$\Rightarrow \frac{1}{\cos \mathit{\theta }}+\frac{1}{\sin \mathit{\theta }}=5$
$\Rightarrow \sin \mathit{\theta }+\cos \mathit{\theta }=5\sin \mathit{\theta }\cos \mathit{\theta }$
$\Rightarrow \sin \mathit{\theta }+\cos \mathit{\theta }=\frac{5}{2}\mathit{s}\mathit{i}\mathit{n}2\mathit{\theta }$
Since sin 2θ or sinθ or cosθ is not given, we can’t determine sinθ + cosθ.
∴ I alone is not sufficient
II. Given : $\sin 2\mathit{\theta }=\frac{1}{2}$
⇒ 2θ = 30° or 150°[Sine is positive in ${\mathit{Q}}_{\mathit{I}}$ and ${\mathit{Q}}_{\mathit{I}\mathit{I}}$]
⇒ θ = 15° or 75°
$\sin 15°+\cos 15°=\sin 15°+\sin 75°$
Also, $\sin 75°+\cos 75°=\sin 75°+\sin 15°$
Hence, $\sin \mathit{\theta }+\cos \mathit{\theta }$ can be uniquely determined.
∴ II alone is sufficient.
Aliter : $\sin \mathit{\theta }+\cos \mathit{\theta }=\sqrt{{\left(\sin \mathit{\theta }+\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\right)}^2}$
$\sqrt{\frac{3}{2}}$
we discard –$\frac{\sqrt{3}}{2}$ as θ lies in the first quadrant.
∴ II alone is sufficient.