====== Question 7, Exercise 8.1 ======
Solutions of Question 7 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 7=====
Given $\alpha$ and $\beta$ are acute angles with $\sin \alpha=\dfrac{12}{13}$ and $\tan \beta=\dfrac{4}{3}$ find \\
(i) $\sin(\alpha+\beta)$ (ii) $\cos(\alpha+\beta)$ (iii) $\tan(\alpha+\beta)$.

** Solution. **

Given: $\sin \alpha=\dfrac{12}{13}$, where $\alpha$ is acute angle, i.e. is in QI.\\
$\tan \beta=\dfrac{4}{3}$, where $\beta$ is acute angle, i.e. is in QI.

We have an identity:
$$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$
As \(\alpha\) lies in QI and \(\cos\) is positive in QI,
\begin{align*}
\cos \alpha & = \sqrt{1-\sin^2\alpha} \\
&= \sqrt{1-{{\left(\frac{12}{13}\right)}^2}} \\
&= \sqrt{1-\dfrac{144}{169}} \\
&= \sqrt{\dfrac{25}{169}} = \dfrac{5}{13}.
\end{align*}

Also,
$$\sec \beta=\pm\sqrt{1+\tan^2\beta}.$$
As \(\beta\) lies in QI and \(\sec\) is positive in QI,
\begin{align*}
\sec \beta & = \sqrt{1+\tan^2\beta} \\
&= \sqrt{1+{{\left(\dfrac{4}{3}\right)}^2}} \\
&= \sqrt{1+\dfrac{16}{9}} \\
&= \sqrt{\dfrac{25}{9}} = \dfrac{5}{3}.
\end{align*}

Thus,
\begin{align*}
\cos \beta & = \frac{1}{\sec \beta} = \frac{3}{5}.
\end{align*}

As
\begin{align*}
\frac{\sin\beta}{\cos\beta} & = \tan\beta \\
\implies \sin\beta & = \tan\beta \cdot \cos\beta \\
&= \left(\frac{4}{3}\right)\left(\frac{3}{5}\right) \\
\implies \sin\beta & = \frac{4}{5}.
\end{align*}

(i) $\sin(\alpha + \beta)$
\begin{align*}
\sin(\alpha + \beta) & = \left(\dfrac{12}{13}\right) \left(\dfrac{3}{5}\right) + \left(\dfrac{5}{13}\right) \left(\dfrac{4}{5}\right) \\
& = \dfrac{36}{65} + \dfrac{20}{65} \\
& = \dfrac{56}{65}.
\end{align*}

(ii) $\cos(\alpha + \beta)$
\begin{align*}
\cos(\alpha + \beta) & = \left(\dfrac{5}{13}\right) \left(\dfrac{3}{5}\right) - \left(\dfrac{12}{13}\right) \left(\dfrac{4}{5}\right) \\
& = \dfrac{15}{65} - \dfrac{48}{65} \\
& = -\dfrac{33}{65}\end{align*}

(ii) $\tan(\alpha + \beta)$
\begin{align*}
\tan(\alpha + \beta) & = \frac{\dfrac{56}{65}}{-\dfrac{33}{65}} \\
& = -\dfrac{56}{33}.
\end{align*}

GOOD
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