Question

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) $\dfrac{\sin {{30}^{0}}17'}{\cos {{59}^{0}}43'}$

Answer 1

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,

& \dfrac{\cos (90-{{37}^{0}})}{\sin {{37}^{0}}} \\\ & =\dfrac{\sin {{37}^{0}}}{\sin {{37}^{0}}} \\\ & =1 \\\ \end{aligned}$$ In question (ii), we have to simplify $$\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{aligned} & \dfrac{tan(90-{{22}^{0}})}{\cot {{22}^{0}}} \\\ & =\dfrac{\cot {{22}^{0}}}{\cot {{22}^{0}}} \\\ & =1 \\\ \end{aligned}$$ In question (iii), we have to simplify $$\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{aligned} & \dfrac{\sec(90-{{41}^{0}})}{\operatorname{\cos ec}{{41}^{0}}} \\\ & =\dfrac{\operatorname{\cos ec}{{41}^{0}}}{\operatorname{\cos ec}{{41}^{0}}} \\\ & =1 \\\ \end{aligned}$$ In question (iv), we have to simplify $$\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{aligned} & \dfrac{\sin {{30}^{0}}17'}{\cos {{59}^{0}}43'} \\\ & =\dfrac{\sin {{30}^{0}}17'}{\sin {{30}^{0}}17'} \\\ & =1 \\\ \end{aligned}$$. Note: In this question, one can make a mistake in writing angles in terms of degree and minutes. One can write $${{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.