Question

Without using trigonometric tables, evaluate:
(i) $\frac{\sin 16°}{\cos 74°}$
(ii) $\frac{\sec 11°}{\mathrm{cosec}79°}$
(iii) $\frac{\tan 27°}{\cot 63°}$
(iv) $\frac{\cos 35°}{\sin 55°}$
(v) $\frac{\mathrm{cosec}42°}{\sec 48°}$
(vi) $\frac{\cot 38°}{\tan 52°}$

Answer 1

$$\begin{gathered} \begin{array}{l}\left(i\right)\frac{sin{16}^{\circ }}{cos{74}^{{}^{\circ }}}\\ \end{array} \\ \begin{array}{l}=\frac{\sin ({90}^{\circ }-{74}^{\circ })}{\cos {74}^{\circ }}\\ \end{array} \\ \begin{array}{l}=\frac{\cos {74}^{\circ }}{\cos {74}^{\circ }}\\ \end{array}[\because \sin \left(90-\theta \right)=\cos \theta ] \\ \begin{array}{l}=1\\ \end{array} \\ \begin{array}{l}\left(\mathrm{ii}\right)\frac{\sec {11}^{\circ }}{\cos \mathrm{ec}{79}^{\circ }}\\ \end{array} \\ \begin{array}{l}=\frac{\sec ({90}^{\circ }-{79}^{\circ })}{\cos {\mathrm{ec79}}^{\circ }}\\ \end{array} \\ \begin{array}{l}=\frac{\cos {\mathrm{ec79}}^{\circ }}{\cos {\mathrm{ec79}}^{\circ }}\\ \end{array}[\because sec\left(90-\theta \right)=\cos ec\theta ] \\ \begin{array}{l}=1\\ \end{array} \\ \begin{array}{l}\left(\mathrm{iii}\right)\frac{\tan {27}^{\circ }}{\cot {63}^{\circ }}\end{array} \\ \begin{array}{l}=\frac{\tan ({90}^{\circ }-{63}^{\circ })}{\cot {63}^{\circ }}\\ \end{array} \\ \begin{array}{l}=\frac{\cot {63}^{\circ }}{\cot {63}^{\circ }}[\because \tan \left(90-\theta \right)=\cot \theta ]\\ \end{array} \\ \begin{array}{l}=1\\ \end{array} \\ \begin{array}{l}\left(\mathrm{iv}\right)\frac{\cos {35}^{\circ }}{\sin {55}^{\circ }}\\ \end{array} \\ \begin{array}{l}=\frac{\cos ({90}^{\circ }-{55}^{\circ })}{\sin {55}^{\circ }}\\ \end{array} \\ \begin{array}{l}=\frac{\sin {55}^{\circ }}{\sin {55}^{\circ }}[\because \sin \left(90-\theta \right)=\cos \theta ]\\ \\ \end{array} \\ =1 \\ \begin{array}{l}\left(v\right)\frac{\cos {\mathrm{ec42}}^{\circ }}{\sec {48}^{\circ }}\\ \end{array} \\ \begin{array}{l}=\frac{\cos \mathrm{ec}({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} \\ \begin{array}{l}\left(\mathrm{vi}\right)\frac{\cot {38}^{\circ }}{\tan {52}^{\circ }}\\ \end{array} \\ \begin{array}{l}=\frac{\cot ({90}^{\circ }-{52}^{\circ })}{\tan {52}^{\circ }}\\ \end{array} \\ \begin{array}{l}=\frac{\tan {52}^{\circ }}{\tan {52}^{\circ }}\\ \end{array}[\because \tan \left(90-\theta \right)=\cot \theta ] \\ \mathrm{=1} \end{gathered}$$