Question 1, Exercise 8.1

Solutions of Question 1 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.

Find the value of $\cos (\alpha \pm \beta), \sin (\alpha \pm \beta)$ and $\tan (\alpha \pm \beta)$ for the pair of angles. $\alpha=180^{\circ}, \beta=60^{\circ}$

Solution.

Given: $\alpha=180^{\circ}$, $\beta=60^{\circ}$.

$$\begin{align*} \cos (\alpha + \beta) & = \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \implies \cos (180+60) & = \cos 180 \cos 60 - \sin 180 \sin 60 \\ \implies \cos (180+60) & = (-1)\left(\frac{1}{2}\right) - (0)\left(\frac{\sqrt{3}}{2}\right) \\ & = -\frac{1}{2} - 0 = -\frac{1}{2} \end{align*}$$

$$\begin{align*} \cos (\alpha - \beta) & = \cos \alpha \cos \beta + \sin \alpha \sin \beta \\ \implies \cos (180-60) & = \cos 180 \cos 60 + \sin 180 \sin 60 \\ \implies \cos (180-60) & = (-1)\left(\frac{1}{2}\right) + (0)\left(\frac{\sqrt{3}}{2}\right) \\ & = -\frac{1}{2} + 0 = -\frac{1}{2} \end{align*}$$

$$\begin{align*} \sin (\alpha + \beta) & = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ \implies \cos (180+60) & = \sin 180 \cos 60 + \cos 180 \sin 60 \\ \implies \cos (180+60) & = (0)\left(\frac{1}{2}\right) + (-1)\left(\frac{\sqrt{3}}{2}\right) \\ & = 0-\frac{\sqrt{3}}{2} + 0 = -\frac{\sqrt{3}}{2} \end{align*}$$

$$\begin{align*} \sin (\alpha - \beta) & = \sin \alpha \cos \beta - \cos \alpha \sin \beta \\ \implies \sin (180-60) & = \sin 180 \cos 60 - \cos 180 \sin 60 \\ \implies\sin (180-60) & = (0)\left(\frac{1}{2}\right) - (-1)\left(\frac{\sqrt{3}}{2}\right) \\ & = 0+\frac{\sqrt{3}}{2} + 0 = \frac{\sqrt{3}}{2} \end{align*}$$

$$\begin{align*} \tan (\alpha + \beta) & = \dfrac{\tan\alpha + \tan\beta}{1-\tan\alpha \tan\beta} \\ \implies \tan (180 + 60) & = \dfrac{\tan 180 + \tan 60}{1-\tan 180 \tan 60} \\ \implies \tan (180 + 60) & = \dfrac{0 + \sqrt{3}}{1-(0)(\sqrt{3})} \\ & = \dfrac{\sqrt{3}}{1-0} = \sqrt{3} \end{align*}$$

$$\begin{align*} \tan (\alpha - \beta) & = \dfrac{\tan\alpha - \tan\beta}{1+\tan\alpha \tan\beta} \\ \implies \tan (180 - 60) & = \dfrac{\tan 180 - \tan 60}{1+\tan 180 \tan 60} \\ \implies \tan (180 - 60) & = \dfrac{0 - \sqrt{3}}{1+(0)(\sqrt{3})} \\ & = \dfrac{-\sqrt{3}}{1+0} = -\sqrt{3} \end{align*}$$

Alternative Method

• $\cos(180+60) = \cos(2(90)+60) = -\cos 60^\circ$
• $\cos(180-60) = \cos(2(90)-60) = -\cos 60^\circ$
• $\sin(180+60) = \sin(2(90)+60) = -\sin 60^\circ$
• $\sin(180-60) = \sin(2(90)-60) = \sin 60^\circ$
• $\tan(180+60) = \tan(2(90)+60) = \tan 60^\circ$
• $\tan(180-60) = \tan(2(90)-60) =-\tan 60^\circ$

Find the value of $\cos (\alpha \pm \beta), \sin (\alpha \pm \beta)$ and $\tan (\alpha \pm \beta)$ for the pair of angles. $\alpha=60^{\circ}, \beta=90^{\circ}$

Solution.

Given: $\alpha = 60^\circ$, $\beta = 90^\circ$ $$\begin{align*} \cos(\alpha + \beta) & = \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \implies\cos(60^\circ + 90^\circ) & = \cos 60^\circ \cos 90^\circ - \sin 60^\circ \sin 90^\circ \\ \implies\cos(60^\circ + 90^\circ) & = \left(\frac{1}{2}\right)(0) - \left(\frac{\sqrt{3}}{2}\right)(1) \\ & = 0 - \frac{\sqrt{3}}{2}\\ \implies cos(60^\circ + 90^\circ) & = -\frac{\sqrt{3}}{2} \end{align*}$$ $$\begin{align*} \cos(\alpha - \beta) & = \cos \alpha \cos \beta + \sin \alpha \sin \beta \\ \implies \cos(60^\circ - 90^\circ) & = \cos 60^\circ \cos 90^\circ + \sin 60^\circ \sin 90^\circ \\ \implies \cos(60^\circ - 90^\circ) & = \left(\frac{1}{2}\right)(0) + \left(\frac{\sqrt{3}}{2}\right)(1) \\ & = 0 + \frac{\sqrt{3}}{2} \\ \implies \cos(60^\circ - 90^\circ) & = \frac{\sqrt{3}}{2} \end{align*}$$ $$\begin{align*} \sin(\alpha + \beta) & = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ \implies \sin(60^\circ + 90^\circ) & = \sin 60^\circ \cos 90^\circ + \cos 60^\circ \sin 90^\circ \\ \implies \sin(60^\circ + 90^\circ) & = \left(\frac{\sqrt{3}}{2}\right)(0) + \left(\frac{1}{2}\right)(1) \\ & = 0 + \frac{1}{2} \\ \implies \sin(60^\circ + 90^\circ) & = \frac{1}{2} \end{align*}$$ $$\begin{align*} \sin(\alpha - \beta) & = \sin \alpha \cos \beta - \cos \alpha \sin \beta \\ \implies \sin(60^\circ - 90^\circ) & = \sin 60^\circ \cos 90^\circ - \cos 60^\circ \sin 90^\circ \\ \implies \sin(60^\circ - 90^\circ) & = \left(\frac{\sqrt{3}}{2}\right)(0) - \left(\frac{1}{2}\right)(1) \\ & = 0 - \frac{1}{2} \\ \implies \sin(60^\circ - 90^\circ)& = -\frac{1}{2} \end{align*}$$ $$\begin{align*} \tan(\alpha + \beta) & = \dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \\ \implies \tan(60^\circ + 90^\circ) & = \dfrac{\tan 60^\circ + \tan 90^\circ}{1 - \tan 60^\circ \tan 90^\circ} \\ \implies \tan(60^\circ + 90^\circ) & = \dfrac{\sqrt{3} + \infty}{1 - (\sqrt{3})(\infty)} \\ & = \sqrt{3} \end{align*}$$ $$\begin{align*} \tan(\alpha - \beta) & = \dfrac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \\ \implies \tan(60^\circ - 90^\circ) & = \dfrac{\tan 60^\circ - \tan 90^\circ}{1 + \tan 60^\circ \tan 90^\circ} \\ \implies \tan(60^\circ - 90^\circ) & = \dfrac{\sqrt{3} - \infty}{1 + (\sqrt{3})(\infty)} \\ & = \sqrt{3} \end{align*}$$

Find the value of $\cos (\alpha \pm \beta), \sin (\alpha \pm \beta)$ and $\tan (\alpha \pm \beta)$ for the pair of angles. $\alpha=180^{\circ}, \beta=30^{\circ}$

Solution.

Given: $\alpha = 180^\circ$, $\beta = 30^\circ$. $$\begin{align*} \cos (\alpha + \beta) & = \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \implies \cos (180^\circ + 30^\circ) & = \cos 180^\circ \cos 30^\circ - \sin 180^\circ \sin 30^\circ \\ \implies \cos (180^\circ + 30^\circ) & = (-1)\left(\frac{\sqrt{3}}{2}\right) - (0)\left(\frac{1}{2}\right) \\ & = -\frac{\sqrt{3}}{2} - 0 \\ & = -\frac{\sqrt{3}}{2} \end{align*}$$ $$\begin{align*} \cos (\alpha - \beta) & = \cos \alpha \cos \beta + \sin \alpha \sin \beta \\ \implies \cos (180^\circ - 30^\circ) & = \cos 180^\circ \cos 30^\circ + \sin 180^\circ \sin 30^\circ \\ \implies \cos (180^\circ - 30^\circ)& = (-1)\left(\frac{\sqrt{3}}{2}\right) + (0)\left(\frac{1}{2}\right) \\ & = -\frac{\sqrt{3}}{2} + 0 \\ & = -\frac{\sqrt{3}}{2} \end{align*}$$ $$\begin{align*} \sin (\alpha + \beta) & = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ \implies \sin (180^\circ + 30^\circ) & = \sin 180^\circ \cos 30^\circ + \cos 180^\circ \sin 30^\circ \\ \implies \cos (180^\circ - 30^\circ) & = (0)\left(\frac{\sqrt{3}}{2}\right) + (-1)\left(\frac{1}{2}\right) \\ & = 0 - \frac{1}{2} \\ & = -\frac{1}{2} \end{align*}$$ $$\begin{align*} \sin (\alpha - \beta) & = \sin \alpha \cos \beta - \cos \alpha \sin \beta \\ \implies \sin (180^\circ - 30^\circ) & = \sin 180^\circ \cos 30^\circ - \cos 180^\circ \sin 30^\circ \\ \implies\sin (180^\circ - 30^\circ) & = (0)\left(\frac{\sqrt{3}}{2}\right) - (-1)\left(\frac{1}{2}\right) \\ & = 0 + \frac{1}{2} \\ & = \frac{1}{2} \end{align*}$$ $$\begin{align*} \tan (\alpha + \beta) & = \dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \\ \implies \tan (180^\circ + 30^\circ) & = \dfrac{\tan 180^\circ + \tan 30^\circ}{1 - \tan 180^\circ \tan 30^\circ} \\ \implies \tan (180^\circ + 30^\circ)& = \dfrac{0 + \frac{1}{\sqrt{3}}}{1 - (0)(\frac{1}{\sqrt{3}})} \\ & = \dfrac{\frac{1}{\sqrt{3}}}{1 - 0} \\ & = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \end{align*}$$ $$\begin{align*} \tan (\alpha - \beta) & = \dfrac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \\ \implies \tan (180^\circ - 30^\circ) & = \dfrac{\tan 180^\circ - \tan 30^\circ}{1 + \tan 180^\circ \tan 30^\circ} \\ \implies \tan (180^\circ - 30^\circ) & = \dfrac{0 - \frac{1}{\sqrt{3}}}{1 + (0)(\frac{1}{\sqrt{3}})} \\ & = \dfrac{-\frac{1}{\sqrt{3}}}{1 + 0} \\ & = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} \end{align*}$$

Find the value of $\cos (\alpha \pm \beta), \sin (\alpha \pm \beta)$ and $\tan (\alpha \pm \beta)$ for the pair of angles. $\alpha=\pi, \beta=\frac{2 \pi}{3}$

Solution.

Given: $\alpha = \pi$, $\beta = \frac{2\pi}{3}$. $$\begin{align*} \cos (\alpha + \beta) & = \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \implies \cos \left( \pi + \frac{2\pi}{3} \right) & = \cos \pi \cos \frac{2\pi}{3} - \sin \pi \sin \frac{2\pi}{3} \\ \implies \cos \left( \pi + \frac{2\pi}{3} \right) & = (-1)\left(-\frac{1}{2}\right) - (0)\left(\frac{\sqrt{3}}{2}\right) \\ & = \frac{1}{2} - 0 \\ & = \frac{1}{2} \end{align*}$$ $$\begin{align*} \cos (\alpha - \beta) & = \cos \alpha \cos \beta + \sin \alpha \sin \beta \\ \implies \cos \left( \pi - \frac{2\pi}{3} \right) & = \cos \pi \cos \frac{2\pi}{3} + \sin \pi \sin \frac{2\pi}{3} \\ \implies \cos \left( \pi - \frac{2\pi}{3} \right) & = (-1)\left(-\frac{1}{2}\right) + (0)\left(\frac{\sqrt{3}}{2}\right) \\ & = \frac{1}{2} + 0 \\ & = \frac{1}{2} \end{align*}$$ $$\begin{align*} \sin (\alpha + \beta) & = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ \implies \sin \left( \pi + \frac{2\pi}{3} \right) & = \sin \pi \cos \frac{2\pi}{3} + \cos \pi \sin \frac{2\pi}{3} \\ \implies \sin \left( \pi + \frac{2\pi}{3} \right) & = (0)\left(-\frac{1}{2}\right) + (-1)\left(\frac{\sqrt{3}}{2}\right) \\ & = 0 - \frac{\sqrt{3}}{2} \\ & = -\frac{\sqrt{3}}{2} \end{align*}$$ $$\begin{align*} \sin (\alpha - \beta) & = \sin \alpha \cos \beta - \cos \alpha \sin \beta \\ \implies \sin \left( \pi - \frac{2\pi}{3} \right) & = \sin \pi \cos \frac{2\pi}{3} - \cos \pi \sin \frac{2\pi}{3} \\ \implies \sin \left( \pi - \frac{2\pi}{3} \right) & = (0)\left(-\frac{1}{2}\right) - (-1)\left(\frac{\sqrt{3}}{2}\right) \\ & = 0 + \frac{\sqrt{3}}{2} \\ & = \frac{\sqrt{3}}{2} \end{align*}$$ $$\begin{align*} \tan (\alpha + \beta) & = \dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \\ \implies \tan \left( \pi + \frac{2\pi}{3} \right) & = \dfrac{\tan \pi + \tan \frac{2\pi}{3}}{1 - \tan \pi \tan \frac{2\pi}{3}} \\ \implies \tan \left( \pi + \frac{2\pi}{3} \right) & = \dfrac{0 + (-\sqrt{3})}{1 - (0)(-\sqrt{3})} \\ & = \dfrac{-\sqrt{3}}{1} \\ & = -\sqrt{3} \end{align*}$$ $$\begin{align*} \tan (\alpha - \beta) & = \dfrac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \\ \implies \tan \left( \pi - \frac{2\pi}{3} \right) & = \dfrac{\tan \pi - \tan \frac{2\pi}{3}}{1 + \tan \pi \tan \frac{2\pi}{3}} \\ \implies \tan \left( \pi - \frac{2\pi}{3} \right) & = \dfrac{0 - (-\sqrt{3})}{1 + (0)(-\sqrt{3})} \\ & = \dfrac{\sqrt{3}}{1} \\ & = \sqrt{3} \end{align*}$$

Find the value of $\cos (\alpha \pm \beta), \sin (\alpha \pm \beta)$ and $\tan (\alpha \pm \beta)$ for the pair of angles. $\alpha=\frac{4 \pi}{3}, \beta=\frac{\pi}{6}$

Solution.

Given: $\alpha = \dfrac{4\pi}{3}$, $\beta = \dfrac{\pi}{6}$.

$$\begin{align*} \cos (\alpha + \beta) & = \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \implies \cos \left( \frac{4\pi}{3} + \frac{\pi}{6} \right) & = \cos \frac{4\pi}{3} \cos \frac{\pi}{6} - \sin \frac{4\pi}{3} \sin \frac{\pi}{6} \\ \implies \cos \left( \frac{3\pi}{2}\right) & = \left(-\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(-\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) \\ & = -\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} \\ & = 0 = \cos \frac{\pi}{2} \end{align*}$$ $$\begin{align*} \cos (\alpha - \beta) & = \cos \alpha \cos \beta + \sin \alpha \sin \beta \\ \implies \cos \left( \frac{4\pi}{3} - \frac{\pi}{6} \right) & = \cos \frac{4\pi}{3} \cos \frac{\pi}{6} + \sin \frac{4\pi}{3} \sin \frac{\pi}{6} \\ \implies \cos \left( \frac{7\pi}{6}\right) & = \left(-\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(-\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) \\ & = -\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} \\ & = -\frac{\sqrt{3}}{2} = \cos \left(\frac{5\pi}{6}\right) \end{align*}$$ $$\begin{align*} \sin (\alpha + \beta) & = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ \implies \sin \left( \frac{4\pi}{3} + \frac{\pi}{6} \right) & = \sin \frac{4\pi}{3} \cos \frac{\pi}{6} + \cos \frac{4\pi}{3} \sin \frac{\pi}{6} \\ \implies \sin \left( \frac{4\pi}{3} + \frac{\pi}{6} \right) & = \left(-\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(-\frac{1}{2}\right)\left(\frac{1}{2}\right) \\ & = -\frac{3}{4} - \frac{1}{4} \\ & = -1 = \sin \left(-\frac{\pi}{2}\right) \end{align*}$$ $$\begin{align*} \sin (\alpha - \beta) & = \sin \alpha \cos \beta - \cos \alpha \sin \beta \\ \implies \sin \left( \frac{4\pi}{3} - \frac{\pi}{6} \right) & = \sin \frac{4\pi}{3} \cos \frac{\pi}{6} - \cos \frac{4\pi}{3} \sin \frac{\pi}{6} \\ \implies \sin \left( \frac{4\pi}{3} - \frac{\pi}{6} \right) & = \left(-\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(-\frac{1}{2}\right)\left(\frac{1}{2}\right) \\ & = -\frac{3}{4} + \frac{1}{4} \\ & = -\frac{1}{2} = \sin \left(-\frac{\pi}{6}\right) \end{align*}$$ $$\begin{align*} \tan (\alpha + \beta) & = \dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \\ \implies \tan \left( \frac{4\pi}{3} + \frac{\pi}{6} \right) & = \dfrac{\tan \frac{4\pi}{3} + \tan \frac{\pi}{6}}{1 - \tan \frac{4\pi}{3} \tan \frac{\pi}{6}} \\ \implies \tan \left( \frac{4\pi}{3} + \frac{\pi}{6} \right) & = \dfrac{-\sqrt{3} + \frac{1}{\sqrt{3}}}{1 - \left(-\sqrt{3}\right)\left(\frac{1}{\sqrt{3}}\right)} \\ & = \dfrac{-\sqrt{3} + \frac{1}{\sqrt{3}}}{1 + 1} \\ & = \dfrac{-\sqrt{3} + \frac{1}{\sqrt{3}}}{2} = \frac{-3 + 1}{2\sqrt{3}} = \frac{-2}{2\sqrt{3}} = \frac{-1}{\sqrt{3}} = -\tan \frac{\pi}{3}\end{align*}$$ $$\begin{align*} \tan (\alpha - \beta) & = \dfrac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \\ \implies \tan \left( \frac{4\pi}{3} - \frac{\pi}{6} \right) & = \dfrac{\tan \frac{4\pi}{3} - \tan \frac{\pi}{6}}{1 + \tan \frac{4\pi}{3} \tan \frac{\pi}{6}} \\ \implies \tan \left( \frac{4\pi}{3} - \frac{\pi}{6} \right) & = \dfrac{-\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \left(-\sqrt{3}\right)\left(\frac{1}{\sqrt{3}}\right)} \\ & = \dfrac{-\sqrt{3} - \frac{1}{\sqrt{3}}}{1 - 1} \quad (\text{undefined}) \end{align*}$$

Find the value of $\cos (\alpha \pm \beta), \sin (\alpha \pm \beta)$ and $\tan (\alpha \pm \beta)$ for the pair of angles. $\alpha=\frac{7 \pi}{4}, \beta=\frac{3 \pi}{4}$

Solution.

Given: $\alpha = \frac{7\pi}{4}$, $\beta = \frac{3\pi}{4}$. $$\begin{align*} \cos (\alpha + \beta) & = \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \implies \cos \left( \frac{7\pi}{4} + \frac{3\pi}{4} \right) & = \cos \frac{7\pi}{4} \cos \frac{3\pi}{4} - \sin \frac{7\pi}{4} \sin \frac{3\pi}{4} \\ \implies \cos (2\pi) & = \left(\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right) - \left(-\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right) \\ & = -\frac{1}{2} + \frac{1}{2} \\ & = 0 = \cos 0 \end{align*}$$ $$\begin{align*} \cos (\alpha - \beta) & = \cos \alpha \cos \beta + \sin \alpha \sin \beta \\ \implies \cos \left( \frac{7\pi}{4} - \frac{3\pi}{4} \right) & = \cos \frac{7\pi}{4} \cos \frac{3\pi}{4} + \sin \frac{7\pi}{4} \sin \frac{3\pi}{4} \\ \implies \cos \left( \pi \right) & = \left(\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right)+\left(-\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right) \\ & = -\frac{1}{2} - \frac{1}{2} \\ & = -1 = \cos \pi \end{align*}$$ $$\begin{align*} \sin (\alpha + \beta) & = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ \implies \sin \left( \frac{7\pi}{4} + \frac{3\pi}{4} \right) & = \sin \frac{7\pi}{4} \cos \frac{3\pi}{4} + \cos \frac{7\pi}{4} \sin \frac{3\pi}{4} \\ \implies \sin (2\pi) & = \left(-\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right) \\ & = \frac{1}{2} + \frac{1}{2} \\ & = 1 = \sin 0 \end{align*}$$ $$\begin{align*} \sin (\alpha - \beta) & = \sin \alpha \cos \beta - \cos \alpha \sin \beta \\ \implies \sin \left( \frac{7\pi}{4} - \frac{3\pi}{4} \right) & = \sin \frac{7\pi}{4} \cos \frac{3\pi}{4} - \cos \frac{7\pi}{4} \sin \frac{3\pi}{4} \\ \implies \sin \left( \pi \right) & = \left(-\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right)-\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right) \\ & = \frac{1}{2} - \frac{1}{2} \\ & = 0 = \sin \pi \end{align*}$$ $$\begin{align*} \tan (\alpha + \beta) & = \dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \\ \implies \tan \left( \frac{7\pi}{4} + \frac{3\pi}{4} \right) & = \dfrac{\tan \frac{7\pi}{4} + \tan \frac{3\pi}{4}}{1 - \tan \frac{7\pi}{4} \tan \frac{3\pi}{4}} \\ \implies \tan (2\pi) & = \dfrac{-1 + (-1)}{1 - (-1)(-1)} \\ & = \dfrac{-2}{1 - 1} \quad (\text{undefined}) \end{align*}$$ $$\begin{align*} \tan (\alpha - \beta) & = \dfrac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \\ \implies \tan \left( \frac{7\pi}{4} - \frac{3\pi}{4} \right) & = \dfrac{\tan \frac{7\pi}{4} - \tan \frac{3\pi}{4}}{1 + \tan \frac{7\pi}{4} \tan \frac{3\pi}{4}} \\ \implies \tan \pi & = \dfrac{-1 - (-1)}{1 + (-1)(-1)} \\ & = \dfrac{0}{1 + 1} \\ & = 0 = \tan \pi \end{align*}$$

Question 2 >