====== Question 9, Exercise 8.1 ======
Solutions of Question 9 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 9(i)=====
Given $\alpha$ and $\beta$ are obtuse angles with $\sin \alpha=\dfrac{1}{\sqrt{2}}$ and $\cos \beta=-\dfrac{3}{5}$ find: $\sin (\alpha \pm \beta)$


** Solution. **

Given: $\sin \alpha=\dfrac{1}{\sqrt{2}}$, $\alpha$ is obtuse angle, i.e. it is in QII.\\
$\cos \beta=-\dfrac{3}{5}$, $\beta$ is obtuse angle, i.e. it is in QII.\\

We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$ 
As $\alpha$ lies in QII and $\cos$ is -ive in QII,
\begin{align*}
\cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ 
&=-\sqrt{1-{{\left(\dfrac{1}{\sqrt{2}}\right)}^{2}}}\\
&=-\sqrt{1-\dfrac{1}{2}}=\sqrt{\dfrac{1}{2}}\\
\implies \cos \alpha&=-\dfrac{1}{\sqrt{2}}.
\end{align*}
Also 
$$\sin\beta=\pm\sqrt{1-\sin^2\beta}.$$
As $\beta$ lies in QII and $\sin$ is +ive in QII,
\begin{align*}
\cos \beta & =\sqrt{1-\cos^2\beta} \\
&=\sqrt{1-{{\left(-\dfrac{3}{5} \right)}^{2}}}\\
&=\sqrt{1-\dfrac{9}{25}} \\
& =\sqrt{\dfrac{16}{25}} =\dfrac{4}{5} 
\end{align*}
\begin{align*}
\sin (\alpha + \beta) &= \sin \alpha \cos \beta + \cos \alpha \sin \beta \\
&= \left( \frac{1}{\sqrt{2}} \right) \left( -\frac{3}{5} \right) + \left( -\frac{1}{\sqrt{2}} \right) \left( \frac{4}{5} \right) \\
&= -\frac{3}{5\sqrt{2}} - \frac{4}{5\sqrt{2}} \\
&= -\frac{7}{5\sqrt{2}}.
\end{align*}
\begin{align*}
\sin (\alpha - \beta) &= \sin \alpha \cos \beta - \cos \alpha \sin \beta \\
&= \left( \frac{1}{\sqrt{2}} \right) \left( -\frac{3}{5} \right) - \left( -\frac{1}{\sqrt{2}} \right) \left( \frac{4}{5} \right) \\
&= -\frac{3}{5\sqrt{2}} + \frac{4}{5\sqrt{2}} \\
&= \frac{1}{5\sqrt{2}}.
\end{align*}

===== Question 9(ii)=====
Given $\alpha$ and $\beta$ are obtuse angles with $\sin \alpha=\frac{1}{\sqrt{2}}$ and $\cos \beta=-\frac{3}{5}$ find: $\cos (\alpha \pm \beta)$

** Solution. **

Given: $\sin \alpha=\dfrac{1}{\sqrt{2}}$, $\alpha$ is obtuse angle, i.e. it is in QII.\\
$\cos \beta=-\dfrac{3}{5}$, $\beta$ is obtuse angle, i.e. it is in QII.\\

We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$ 
As $\alpha$ lies in QII and $\cos$ is -ive in QII,
\begin{align*}
\cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ 
&=-\sqrt{1-{{\left(\dfrac{1}{\sqrt{2}}\right)}^{2}}}\\
&=-\sqrt{1-\dfrac{1}{2}}=\sqrt{\dfrac{1}{2}}\\
\implies \cos \alpha&=-\dfrac{1}{\sqrt{2}}.
\end{align*}
Also 
$$\sin\beta=\pm\sqrt{1-\sin^2\beta}.$$
As $\beta$ lies in QII and $\sin$ is +ive in QII,
\begin{align*}
\cos \beta & =\sqrt{1-\cos^2\beta} \\
&=\sqrt{1-{{\left(-\dfrac{3}{5} \right)}^{2}}}\\
&=\sqrt{1-\dfrac{9}{25}} \\
& =\sqrt{\dfrac{16}{25}} =\dfrac{4}{5} 
\end{align*}
\begin{align*}
\cos (\alpha + \beta) &= \left( -\frac{1}{\sqrt{2}} \right) \left( -\frac{3}{5} \right) - \left( \frac{1}{\sqrt{2}} \right) \left( \frac{4}{5} \right) \\
&= \frac{3}{5\sqrt{2}} - \frac{4}{5\sqrt{2}} \\
&= -\frac{1}{5\sqrt{2}}.
\end{align*}
\begin{align*}
\cos (\alpha - \beta) &= \left( -\frac{1}{\sqrt{2}} \right) \left( -\frac{3}{5} \right) + \left( \frac{1}{\sqrt{2}} \right) \left( \frac{4}{5} \right) \\
&= \frac{3}{5\sqrt{2}} + \frac{4}{5\sqrt{2}} \\
&= \frac{7}{5\sqrt{2}}.
\end{align*}


===== Question 9(iii)=====
Given $\alpha$ and $\beta$ are obtuse angles with $\sin \alpha=\frac{1}{\sqrt{2}}$ and $\cos \beta=-\frac{3}{5}$ find: $\tan (\alpha \pm \beta)$

** Solution. **

Given: $\sin \alpha=\dfrac{1}{\sqrt{2}}$, $\alpha$ is obtuse angle, i.e. it is in QII.\\
$\cos \beta=-\dfrac{3}{5}$, $\beta$ is obtuse angle, i.e. it is in QII.\\

We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$ 
As $\alpha$ lies in QII and $\cos$ is -ive in QII,
\begin{align*}
\cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ 
&=-\sqrt{1-{{\left(\dfrac{1}{\sqrt{2}}\right)}^{2}}}\\
&=-\sqrt{1-\dfrac{1}{2}}\\
\implies \cos \alpha&=-\dfrac{1}{\sqrt{2}}.
\end{align*}
Now \begin{align*}
tan \alpha&=\frac{\sin \alpha}{\cos \alpha}\\
&=\frac{\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{\sqrt{2}}}\\
tan \alpha&= -1
\end{align*}
As $\beta$ lies in QII and $\sin$ is +ive in QII,
\begin{align*}
\sin \beta & =\sqrt{1-\cos^2\beta} \\
&=\sqrt{1-{{\left(-\dfrac{3}{5} \right)}^{2}}}\\
&=\sqrt{1-\dfrac{9}{25}} \\
& =\sqrt{\dfrac{16}{25}} =\dfrac{4}{5} 
\end{align*}
Now \begin{align*}
tan \beta&=\frac{\sin \beta}{\cos \beta}\\
&=\frac{\dfrac{4}{5}}{-\dfrac{3}{5}}\\
tan \beta &= -\dfrac{4}{3}
\end{align*}
\begin{align*}
\tan(\alpha + \beta) &= \frac{-1 + \left( -\frac{4}{3} \right)}{1 - \left( -1 \right) \left( -\frac{4}{3} \right)} \\
&= \frac{-1 - \frac{4}{3}}{1 - \frac{4}{3}} \\
&= \frac{-\frac{7}{3}}{\frac{-1}{3}} \\
&= 7.
\end{align*}
\begin{align*}
\tan(\alpha - \beta) &= \frac{-1 - \left( -\frac{4}{3} \right)}{1 + \left( -1 \right) \left( -\frac{4}{3} \right)} \\
&= \frac{-1 + \frac{4}{3}}{1 + \frac{4}{3}} \\
&= \frac{\frac{1}{3}}{\frac{7}{3}} \\
&= \frac{1}{7}.
\end{align*}



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