Question 8, Exercise 8.1

Solutions of Question 8 of Exercise 8.1 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

If $\sin \alpha=\dfrac{3}{5}$, where $0<\alpha<\dfrac{\pi}{2}$ and $\cos \beta=\dfrac{12}{13}$, where $\dfrac{3 \pi}{2}<\beta<2 \pi$ find:
(i) $\csc (\alpha+\beta)$ (ii) $\sec (\alpha+\beta)$ (iii) $\cot (\alpha+\beta)$

Solution.

Given: $\sin \alpha=\dfrac{3}{5}$, where $0<\alpha<\dfrac{\pi}{2}$, i.e. $\alpha$ lies in QI.
\begin{align*} \cos \alpha &= \sqrt{1 - \sin^2 \alpha} \\ &= \sqrt{1 - \left( \frac{3}{5} \right)^2} \\ &= \sqrt{\frac{16}{25}}\\ \cos \alpha & = \frac{4}{5}. \end{align*} $\cos \beta=\dfrac{12}{13}$, where $\dfrac{3 \pi}{2}<\beta<2 \pi$, i.e. $\beta$ lies in QIV. \begin{align*} \sin \beta &= -\sqrt{1 - \cos^2 \beta} \\ &= -\sqrt{1 - \left( \frac{12}{13} \right)^2} \\ &= -\sqrt{\frac{25}{169}}\\ \sin \beta & = -\frac{5}{13}. \end{align*} (i) \begin{align*} \sin (\alpha + \beta) &= \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ &= \left( \frac{3}{5} \right) \left( \frac{12}{13} \right) + \left( \frac{4}{5} \right) \left( -\frac{5}{13} \right) \\ &= \frac{36}{65} - \frac{20}{65} \\ &= \frac{16}{65}. \end{align*} \begin{align*} \csc (\alpha + \beta) &= \frac{1}{\sin (\alpha + \beta)} \csc (\alpha + \beta) &= \frac{1}{\dfrac{16}{65}} \\ &= \frac{65}{16}. \end{align*} (ii) \begin{align*} \cos (\alpha + \beta) &= \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ &= \left( \frac{4}{5} \right) \left( \frac{12}{13} \right) - \left( \frac{3}{5} \right) \left( -\frac{5}{13} \right) \\ &= \frac{48}{65} + \frac{15}{65} \\ &= \frac{63}{65}. \end{align*} \begin{align*} \sec (\alpha + \beta) &= \frac{1}{\cos (\alpha + \beta)} \sec (\alpha + \beta) &= \frac{1}{\dfrac{63}{65}} \\ &= \frac{65}{63}. \end{align*} (iii) \begin{align*} \cot (\alpha + \beta) &= \frac{\cos (\alpha + \beta)}{\sin (\alpha + \beta)} \cot (\alpha + \beta) &= \frac{\dfrac{63}{65}}{\dfrac{16}{65}} \\ &= \frac{63}{16}. \end{align*}