====== Question 5 and 6, Exercise 8.1 ======
Solutions of Question 5 and 6 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 5=====
For $\sin \alpha=\dfrac{4}{5}, \tan \beta=-\dfrac{5}{12}$ with terminal side of an angles in QII, find $\cos (\alpha+\beta)$ and $\cos (\alpha-\beta)$.

** Solution. **

Given: $\sin \alpha=\dfrac{4}{5}$, $\alpha$ is in QII and $\tan \beta=-\dfrac{5}{12}$, $\beta$ 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(-\frac{4}{5} \right)}^{2}}}\\
&=-\sqrt{1-\dfrac{16}{25}}=\sqrt{\dfrac{9}{25}}\\
\Rightarrow \quad \cos \alpha&=-\dfrac{3}{5}.\end{align*}
Also $$\sec \beta=\pm\sqrt{1+\tan^2\beta}.$$
As $\beta$ lies in QII and $\sec$ is -ive in QII,
\begin{align*}\sec \beta & =-\sqrt{1+\tan^2\beta} \\
&=-\sqrt{1+{{\left(-\dfrac{5}{12} \right)}^{2}}}\\
&=-\sqrt{1+\dfrac{25}{144}} \\
& =-\sqrt{\dfrac{169}{144}} =-\dfrac{13}{12} 
\end{align*}

\begin{align*} \Rightarrow \quad \cos \beta=\frac{1}{\sec \beta}=-\frac{12}{13}.\end{align*}
This gives
\begin{align*}
\frac{\sin\beta}{\cos\beta} & = \tan\beta \\
\implies  \sin\beta & = \tan\beta \cos\beta \\
& = \left(-\frac{5}{12} \right) \left(-\frac{12}{13} \right) \\
\implies  \sin\beta & = \frac{5}{13}.
\end{align*}
Now
\begin{align*}\cos (\alpha+\beta )&=\cos \alpha\cos \beta-\sin \alpha\sin \beta \\ 
&=\left(-\frac{3}{5}\right)\left(-\frac{12}{13}\right)-\left(\frac{4}{5}\right)\left(\frac{5}{13}\right)\\
&=\dfrac{36}{65}-\frac{20}{65} =\frac{16}{65} \end{align*} 
Also
\begin{align*}\cos (\alpha-\beta )&=\cos \alpha\cos \beta+\sin \alpha\sin \beta \\ 
&=\left(-\frac{3}{5}\right)\left(-\frac{12}{13}\right)+\left(\frac{4}{5}\right)\left(\frac{5}{13}\right)\\
&=\dfrac{36}{65}+\frac{20}{65} =\frac{56}{65} \end{align*} 
Hence
\begin{align*}\boxed{\cos(\alpha+\beta)=\frac{16}{65}} \text{ and } \boxed{\cos(\alpha-\beta)=\frac{56}{65}}\end{align*} 
  
===== Question 6=====
For $\cos \alpha=-\dfrac{7}{25}$ with terminal side of $\alpha$ in QII and $\cot \beta=\dfrac{15}{8}$ with terminal side of $\beta$ in QIII, find \\
(i) $\sin(\alpha-\beta)$ (ii) $\cos(\alpha-\beta)$ (iii) $\tan(\alpha-\beta)$.

** Solution. **

Given: $\cos\alpha=-\dfrac{7}{25}$, $\alpha$ is in QII and $\cot\beta=\dfrac{15}{8}$, $\beta$ is in QIII.

We have an identity: $$\sin \alpha=\pm \sqrt{1-\cos^2\alpha}.$$ 
As $\alpha$ lies in QII and $\sin$ is +ive in QII,
\begin{align*}\sin\alpha &=\sqrt{1-\cos^2\alpha}\\ 
&=\sqrt{1-{{\left(-\frac{7}{25} \right)}^{2}}}\\
&=\sqrt{1-\dfrac{49}{625}}=\sqrt{\dfrac{576}{625}}\\
\implies \sin \alpha&=\dfrac{24}{25}.\end{align*}
Also $$\csc \beta=\pm\sqrt{1+\cot^2\beta}.$$
As $\beta$ lies in QIII and $\csc$ is -ive in QIII,
\begin{align*}\csc \beta & =-\sqrt{1+\cot^2\beta} \\
&=-\sqrt{1+{{\left(-\dfrac{15}{8} \right)}^{2}}}\\
&=-\sqrt{1+\dfrac{225}{64}} \\
& =-\sqrt{\dfrac{289}{64}} =-\dfrac{17}{8} 
\end{align*}

\begin{align*} \Rightarrow \quad \sin \beta=\frac{1}{\csc \beta}=-\frac{8}{17}.\end{align*}
This gives
\begin{align*}
\frac{\cos\beta}{\sin\beta} & = \cot\beta \\
\implies  \cos\beta & = \cot\beta \sin\beta \\
& = \left(\frac{15}{8} \right) \left(-\frac{8}{17} \right) \\
\implies  \cos\beta & = -\frac{15}{17}.
\end{align*}

(i) $\sin(\alpha-\beta)$
\begin{align*}\sin (\alpha-\beta )&=\sin \alpha\cos \beta-\cos \alpha\sin \beta \\ 
&=\left(\frac{24}{25}\right)\left(-\frac{15}{17}\right)-\left(-\frac{7}{25}\right)\left(-\frac{8}{17}\right)\\
&=-\dfrac{72}{85}-\frac{56}{425} =-\frac{416}{425} \end{align*} 

(ii) $\cos(\alpha-\beta)$
\begin{align*}\cos (\alpha-\beta )&=\cos \alpha\cos \beta+\sin \alpha\sin \beta \\ 
&=\left(-\frac{7}{25}\right)\left(-\frac{15}{17}\right)+\left(\frac{24}{25}\right)\left(-\frac{8}{17}\right)\\
&=\dfrac{21}{85}-\frac{192}{425} =-\frac{87}{425} \end{align*}

(iii) $\tan(\alpha-\beta)$
\begin{align*}\tan(\alpha-\beta) & =  \frac{\sin(\alpha-\beta)}{\cos (\alpha-\beta)}\\
&= \frac{-416/425}{-87/425} = \frac{416}{87}
\end{align*}

Hence \\
$\boxed{\sin(\alpha-\beta)=-\frac{416}{425}}$, $\boxed{\cos(\alpha-\beta)=-\frac{87}{425}}$, $\boxed{\tan(\alpha-\beta)=\frac{416}{87}}$.
GOOD

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