Correct Answer – Option 3 : -sec 18° 

Concept:

Trigonometric Formulas:

\(\ \rm cosec \;\theta = \dfrac 1{\sin \;\theta}\)

 \(\sin \;(-\theta )= -\sin\;\theta\)

sin (A – B) = sin A. sin B – cos A cos B

Calculations:

Consider,  \(\left(\dfrac{\sec 18^\circ}{\sec 144^\circ} + \dfrac{\text{cosec} \ 18^\circ}{\text{cosec} \ 144^\circ}\right)\) 

=\(\left(\dfrac{ \dfrac 1 {\cos 18^\circ}}{\dfrac 1 {\cos 144^\circ}} + \dfrac{\dfrac 1 {\sin\ 18^\circ}}{\dfrac 1 {\sin \ 144^\circ}}\right)\)

=\(\left(\dfrac{\cos 144^\circ}{\cos 18^\circ} + \dfrac{\text{sin} \ 144^\circ}{\text{sin} \ 18^\circ}\right)\)

=\(\left(\dfrac{\cos (180^\circ – 36^\circ)}{\cos 18^\circ} + \dfrac{\text{sin}( \ 180^\circ-36^\circ)}{\text{sin} \ 18^\circ}\right)\)

=\(\left(\dfrac{\cos 36^\circ}{\cos 18^\circ} – \dfrac{\text{sin}36^\circ}{\text{sin} \ 18^\circ}\right)\)

=\(\left(\dfrac{\cos 36^\circ \sin 18^\circ -{\sin}36^\circ\cos 18^\circ }{\text{sin} \ 18^\circ.\cos 18^\circ}\right)\)

=\(\left(\dfrac{\sin (18^\circ -36^\circ) }{\text{sin} \ 18^\circ.\cos 18^\circ}\right)\)

=\(-\left(\dfrac{\sin 18^\circ }{\text{sin} \ 18^\circ.\cos 18^\circ}\right)\)

=\(-\left(\dfrac{1 }{\cos 18^\circ}\right)\)

= -sec 18° 

Hence,  \(\left(\dfrac{\sec 18^\circ}{\sec 144^\circ} + \dfrac{\text{cosec} \ 18^\circ}{\text{cosec} \ 144^\circ}\right) =-\sec 18^\circ\)