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.

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

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