Question

Without using trigonometric tables, evaluate:
(i) $\frac{\sin 26°}{\cos 64°}$

(ii) $\frac{\sec 11°}{\mathrm{cosec}79°}$

(iii) $\frac{\tan 65°}{\cot 25°}$

(iv) $\frac{\cos 37°}{\sin 53°}$

(v) $\frac{\mathrm{cosec}42°}{\sec 48°}$

(vi) $\frac{\cot 34°}{\tan 56°}$

Answer 1

(i) $\frac{\sin 26°}{\cos 64°}$


$$\begin{gathered} \frac{\sin 26°}{\cos 64°}=\frac{\sin \left(90°-64°\right)}{\cos 64°} \\ =\frac{\cos 64°}{\cos 64°}\left(\because \sin \left(90°-\theta \right)=\cos \theta \right) \\ =1 \\ \mathrm{Hence},\frac{\sin 26°}{\cos 64°}=1 \end{gathered}$$

$$\begin{gathered} \begin{array}{l}\left(\mathrm{ii}\right)\frac{\sec {11}^{\circ }}{\mathrm{cosec}{79}^{\circ }}\\ \end{array} \\ \begin{array}{l}=\frac{\sec ({90}^{\circ }-{79}^{\circ })}{\mathrm{cosec}{79}^{\circ }}\\ \end{array} \\ \begin{array}{l}=\frac{\mathrm{cosec}{79}^{\circ }}{\mathrm{cosec}{79}^{\circ }}\\ \end{array}[\because sec\left(90-\theta \right)=\cos ec\theta ] \\ \begin{array}{l}=1\\ \end{array} \end{gathered}$$


(iii) $\frac{\tan 65°}{\cot 25°}$


$$\begin{gathered} \frac{\tan 65°}{\cot 25°}=\frac{\tan \left(90°-25°\right)}{\cot 25°} \\ =\frac{\cot 25°}{\cot 25°}\left(\because \tan \left(90°-\theta \right)=\cot \theta \right) \\ =1 \\ \mathrm{Hence},\frac{\tan 65°}{\cot 25°}=1 \end{gathered}$$


(iv) $\frac{\cos 37°}{\sin 53°}$


$$\begin{gathered} \frac{\cos 37°}{\sin 53°}=\frac{\cos \left(90°-53°\right)}{\sin 53°} \\ =\frac{\sin 53°}{\sin 53°}\left(\because \cos \left(90°-\theta \right)=\sin \theta \right) \\ =1 \\ \mathrm{Hence},\frac{\cos 37°}{\sin 53°}=1 \end{gathered}$$

$$\begin{gathered} \begin{array}{l}\left(v\right)\frac{\mathrm{cosec}{42}^{\circ }}{\sec {48}^{\circ }}\\ \end{array} \\ \begin{array}{l}=\frac{\mathrm{cosec}({90}^{\circ }-{48}^{\circ })}{\sec {48}^{\circ }}\\ \end{array} \\ \begin{array}{l}=\frac{\sec {48}^{\circ }}{\sec {48}^{\circ }}[\because sec\left(90-\theta \right)=\cos ec\theta ]\\ \end{array} \\ \begin{array}{l}=1\\ \end{array} \end{gathered}$$


(vi) $\frac{\cot 34°}{\tan 56°}$



$$\begin{gathered} \frac{\cot 34°}{\tan 56°}=\frac{\cot \left(90°-56°\right)}{\tan 56°} \\ =\frac{\tan 56°}{\tan 56°}\left(\because \cot \left(90°-\theta \right)=\tan \theta \right) \\ =1 \\ \mathrm{Hence},\frac{\cot 34°}{\tan 56°}=1 \end{gathered}$$