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# What is $(\dfrac{{\sec 18^\circ }}{{\sec 144^\circ }} + \dfrac{{{\text{cosec18}}^\circ }}{{{\text{cosec144}}^\circ }})$ equal to?A) ${\text{sec18}}^\circ$B) ${\text{cosec18}}^\circ$C) ${\text{ - sec18}}^\circ$D) ${\text{ - cosec18}}^\circ$

Last updated date: 22nd Jun 2024
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Hint: Here we can see secant and cosecant are the reciprocals of cosine and sine respectively. So we can replace them. Then we can replace the angle $144$ by $180 - 36$. So we can apply equations of $\cos 2\theta$ and $\sin 2\theta$. Then solving using trigonometric relations we get the answer.

Formula used: For any angle $\theta$ we have the following trigonometric relations.
$\sec \theta = \dfrac{1}{{\cos \theta }}$ and ${\text{cosec}}\theta = \dfrac{1}{{\sin \theta }}$
$\cos (180 - \theta ) = - \cos \theta$ and $\sin (180 - \theta ) = \sin \theta$
$\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta$
$\sin 2\theta = 2\sin \theta \cos \theta$
${\sin ^2}\theta + {\cos ^2}\theta = 1$

Complete step-by-step solution:
We have to find the value of $(\dfrac{{\sec 18^\circ }}{{\sec 144^\circ }} + \dfrac{{{\text{cosec18}}^\circ }}{{{\text{cosec144}}^\circ }})$.
Since $\sec \theta = \dfrac{1}{{\cos \theta }}$ and ${\text{cosec}}\theta = \dfrac{1}{{\sin \theta }}$, we can write
$\Rightarrow$$\dfrac{{\sec 18^\circ }}{{\sec 144^\circ }} + \dfrac{{{\text{cosec18}}^\circ }}{{{\text{cosec144}}^\circ }} = \dfrac{{\cos 144^\circ }}{{\cos 18^\circ }} + \dfrac{{{\text{sin144}}^\circ }}{{{\text{sin18}}^\circ }} We can replace $144$ by 180 - 36. So we have, \Rightarrow$$\dfrac{{\sec 18^\circ }}{{\sec 144^\circ }} + \dfrac{{{\text{cosec18}}^\circ }}{{{\text{cosec144}}^\circ }} = \dfrac{{\cos (180 - 36)^\circ }}{{\cos 18^\circ }} + \dfrac{{{\text{sin}}(180 - 36)^\circ }}{{{\text{sin18}}^\circ }}$
For angles less than ${90^ \circ }$, $180 - \theta$ belongs to the second quadrant.
In the second quadrant sine and cosine values are positive and all other values are negative.
We know $\cos (180 - \theta ) = - \cos \theta$ and $\sin (180 - \theta ) = \sin \theta$
So we get from the above equation,
$\Rightarrow$$\dfrac{{\sec 18^\circ }}{{\sec 144^\circ }} + \dfrac{{{\text{cosec18}}^\circ }}{{{\text{cosec144}}^\circ }} = - \dfrac{{\cos 36^\circ }}{{\cos 18^\circ }} + \dfrac{{{\text{sin}}36^\circ }}{{{\text{sin18}}^\circ }} Now we have the equations: \cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta \sin 2\theta = 2\sin \theta \cos \theta Using these equations we can write, \Rightarrow$$\cos 36^\circ = {\cos ^2}18^\circ - {\sin ^2}18^\circ$
$\Rightarrow$$\sin 36^\circ = 2\sin 18^\circ \cos 18^\circ Substituting these in the above equation we have, \Rightarrow$$\dfrac{{\sec 18^\circ }}{{\sec 144^\circ }} + \dfrac{{{\text{cosec18}}^\circ }}{{{\text{cosec144}}^\circ }} = - (\dfrac{{{{\cos }^2}18^\circ - {{\sin }^2}18^\circ }}{{\cos 18^\circ }}) + \dfrac{{2\sin 18^\circ \cos 18^\circ }}{{{\text{sin18}}^\circ }}$
Cancelling $\sin 18^\circ$ from numerator and denominator on the second term of right hand side of the above equation,
We have,
$\Rightarrow$$\dfrac{{\sec 18^\circ }}{{\sec 144^\circ }} + \dfrac{{{\text{cosec18}}^\circ }}{{{\text{cosec144}}^\circ }} = \dfrac{{{{\sin }^2}18^\circ - {{\cos }^2}18^\circ }}{{\cos 18^\circ }} + 2\cos 18^\circ Simplifying the above equation we get, \Rightarrow$$\dfrac{{\sec 18^\circ }}{{\sec 144^\circ }} + \dfrac{{{\text{cosec18}}^\circ }}{{{\text{cosec144}}^\circ }} = \dfrac{{{{\sin }^2}18^\circ - {{\cos }^2}18^\circ + 2{{\cos }^2}18^\circ }}{{\cos 18^\circ }}$
$\Rightarrow \dfrac{{\sec 18^\circ }}{{\sec 144^\circ }} + \dfrac{{{\text{cosec18}}^\circ }}{{{\text{cosec144}}^\circ }} = \dfrac{{{{\sin }^2}18^\circ + {{\cos }^2}18^\circ }}{{\cos 18^\circ }}$
Also we know that
$sin^2 \theta + cos^2 \theta =1$
Using this result in the above equation we get,
$\Rightarrow$$\dfrac{{\sec 18^\circ }}{{\sec 144^\circ }} + \dfrac{{{\text{cosec18}}^\circ }}{{{\text{cosec144}}^\circ }} = \dfrac{1}{{\cos 18^\circ }} \sec \theta = \dfrac{1}{{\cos \theta }} \Rightarrow$$\dfrac{{\sec 18^\circ }}{{\sec 144^\circ }} + \dfrac{{{\text{cosec18}}^\circ }}{{{\text{cosec144}}^\circ }} = \sec 18^\circ$
Thus we get the solution.

$\therefore$ The answer is option A.

Note: We changed the trigonometric ratios to $\sin$ and $\cos$ so that we can simplify easily. There are different trigonometric rules to solve these types of problems. We have to choose the appropriate equations in each step. For $\cos 2\theta$ we have four equations. But here we chose this one so that we can simplify the answer.