Without using trigonometric tables, evaluate:
(i) \[\dfrac{\cos {{53}^{0}}}{\sin {{37}^{0}}}\]
(ii) \[\dfrac{\tan {{68}^{0}}}{\cot {{22}^{0}}}\]
(iii) \[\dfrac{sec4{{9}^{0}}}{\cos ec{{41}^{0}}}\]
(iv) \[\dfrac{\sin {{30}^{0}}17'}{\cos {{59}^{0}}43'}\]

Hint: For this question, we know the trigonometric identities which are \[\cos (90-\theta )=sin\theta \] ,\[\tan (90-\theta )=\cot\theta \] , and \[\sec (90-\theta )=\cos ec\theta \] . Now, in question (i) using the identity \[\cos (90-\theta )=\sin\theta \] , where \[\theta ={{37}^{0}}\]. Similarly, in question (ii) and question (iii), we have to use \[\tan (90-\theta )=\cot\theta \] , \[\sec (90-\theta )=\cos ec\theta \] , where \[\theta ={{22}^{0}}\] and \[\theta ={{41}^{0}}\] respectively. In question (iv), we have to use the identity \[\cos (90-\theta )=\sin\theta \] , where \[\theta ={{30}^{0}}17'\] . We can write \[{{90}^{0}}\] as \[{{89}^{0}}60'\] .

Complete step-by-step solution -
In question (i), we have to simplify \[\dfrac{\cos {{53}^{0}}}{\sin {{37}^{0}}}\] ……………..(1)
We know the identity, \[\cos (90-\theta )=\sin\theta \] ………….(2)
Replacing \[\theta \] by \[{{37}^{0}}\] in equation (2), we get
\[\cos{{53}^{0}}=\cos (90-{{37}^{0}})=\sin{{37}^{0}}\] ………………..(3)
Now, using equation (3) we can transform equation (1) as,
\[\begin{align}
  & \dfrac{\cos (90-{{37}^{0}})}{\sin {{37}^{0}}} \\
 & =\dfrac{\sin {{37}^{0}}}{\sin {{37}^{0}}} \\
 & =1 \\
\end{align}\]
 In question (ii), we have to simplify \[\dfrac{\tan {{68}^{0}}}{\cot {{22}^{0}}}\] ……………..(4)
We know the identity, \[\tan (90-\theta )=\cot\theta \]………….(5)
Replacing \[\theta \] by \[{{22}^{0}}\] in equation (6), we get
\[\tan{{68}^{0}}=\tan(90-{{22}^{0}})=\cot {{22}^{0}}\] ………………..(6)
Now, using equation (6) we can transform equation (4) as,
\[\begin{align}
  & \dfrac{tan(90-{{22}^{0}})}{\cot {{22}^{0}}} \\
 & =\dfrac{\cot {{22}^{0}}}{\cot {{22}^{0}}} \\
 & =1 \\
\end{align}\]
In question (iii), we have to simplify \[\dfrac{\sec4{{9}^{0}}}{\cos ec{{41}^{0}}}\] ……………..(7)
We know the identity, \[\cos (90-\theta )=\sin\theta \] ………….(8)
Replacing \[\theta \] by \[{{41}^{0}}\] in equation (8), we get
\[\sec{{49}^{0}}=\sec (90-{{41}^{0}})=\cos ec{{41}^{0}}\] ………………..(9)
Now, using equation (9) we can transform equation (7) as,
\[\begin{align}
  & \dfrac{\sec(90-{{41}^{0}})}{\operatorname{\cos ec}{{41}^{0}}} \\
 & =\dfrac{\operatorname{\cos ec}{{41}^{0}}}{\operatorname{\cos ec}{{41}^{0}}} \\
 & =1 \\
\end{align}\]
In question (iv), we have to simplify \[\dfrac{\sin {{30}^{0}}17'}{\cos {{59}^{0}}43'}\] ……………..(10)
We know the identity, \[\cos (90-\theta )=\sin\theta \] and $1^\circ = 60’$………….(11)
We can write \[{{90}^{0}}\] as \[{{89}^{0}}60'\] .
Replacing \[\theta \] by \[{{30}^{0}}17'\] , we get
\[\cos{{59}^{0}}43' = \cos ({{89}^{0}}60'-{{30}^{0}}17')=\sin{{30}^{0}}17'\] ………………..(12)
Now, using equation (12) we can transform equation (10) as,
\[\begin{align}
  & \dfrac{\sin {{30}^{0}}17'}{\cos {{59}^{0}}43'} \\
 & =\dfrac{\sin {{30}^{0}}17'}{\sin {{30}^{0}}17'} \\
 & =1 \\
\end{align}\].

Note: In this question, one can make a mistake in writing angles in terms of degree and minutes. One can write \[{{90}^{0}}\] as \[{{90}^{0}}60'\] . This is wrong. If we write \[{{90}^{0}}60'\] , it means \[{{91}^{0}}\]. So, keep in mind that one degree is 60 minutes.