NIMCET 2017 Question Paper with solutions

Concept:
$\cos \left(\mathit{A}+\mathit{B}\right)=\cos \mathit{A}\cos \mathit{B}-\sin \mathit{A}\sin \mathit{B}$
${\sin }^2\mathit{x}+{\cos }^2\mathit{x}=1$ In the fourth quadrant, the values for cos are positive only.
Calculation:
${\sin }^2\mathit{\theta }=1-{\cos }^2\mathit{\theta }=1-{\left(\frac{4}{5}\right)}^2=\frac{9}{25}$
$\sin \mathit{\theta }=\pm 3∕5$
$=-3∕5$
${\sin }^2\mathit{\phi }=1-{\cos }^2\mathit{\phi }=1-{\left(\frac{12}{13}\right)}^2=\frac{25}{169}$
$\sin \mathit{\phi }=\pm 5∕13$
$=-5∕13$

$=\frac{4}{5}\times \frac{12}{13}\times -\left(\frac{-3}{5}\times \frac{-5}{13}\right)=\frac{33}{65}$
Hence, option (3) is correct.