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.
Question 8
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*}
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