====== Question, Exercise 10.1======

Solutions of Question 4 of Exercise 10.1 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
=====Question 4(i)===== 
If $\sin \alpha =-\dfrac{4}{5}$ and $\cos \beta =-\dfrac{12}{13}$,  $\alpha $in Quadrant III  and $\beta $in Quadrant II, find the exact value of $\sin \left( \alpha -\beta  \right)$.
====Solution====
Given: $\sin \alpha=-\dfrac{4}{5}$, $\alpha$ is in 3rd quadrant, \\
 $\sin \beta=-\dfrac{12}{13}$, $\beta$ is in 2nd quadrant.
We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$ 
As $\alpha$ lies in 3rd quadrant and $\cos$ is -ive in 3rd quadrant,
\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 $$\sin \beta=\pm\sqrt{1-\cos^2\beta}.$$
As $\beta$ lies in 2nd quadrant and $\sin$ is +ive in 2nd quadrant,
\begin{align}&=\sqrt{1-{{\left(-\dfrac{12}{13} \right)}^{2}}}\\
&=\sqrt{1-\dfrac{144}{169}}=\sqrt{\dfrac{25}{169}}\\
\Rightarrow \quad \sin \beta &=\frac{5}{13}.\end{align}
Now 
\begin{align}\sin (\alpha-\beta )&=\sin \alpha\cos \beta+\cos \alpha\sin \beta \\ 
&=\left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right)+\left(-\frac{3}{5}\right)\left(\frac{5}{13}\right)\\
&=\dfrac{48}{65}-\frac{15}{65}\\
\Rightarrow \quad \cos \left( \alpha+\beta \right)&=\frac{33}{65}.\end{align} 

=====Question 4(ii)=====
If $\sin \alpha =-\dfrac{4}{5}$ and $\cos \beta =-\dfrac{12}{13}$,  $\alpha $in Quadrant III  and $\beta $in Quadrant II, find the exact value of $\cos \left( \alpha +\beta  \right)$.
====Solution====
Given: $\sin \alpha=-\dfrac{4}{5}$, $\alpha$ is in 3rd quadrant, \\
 $\sin \beta=-\dfrac{12}{13}$, $\beta$ is in 2nd quadrant.

We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$ 
As $\alpha$ lies in 3rd quadrant and $\cos$ is -ive in 3rd quadrant,
\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 $$\sin \beta=\pm\sqrt{1-\cos^2\beta}.$$
As $\beta$ lies in 2nd quadrant and $\sin$ is +ive in 2nd quadrant,
\begin{align}&=\sqrt{1-{{\left(-\dfrac{12}{13} \right)}^{2}}}\\
&=\sqrt{1-\dfrac{144}{169}}=\sqrt{\dfrac{25}{169}}\\
\Rightarrow \quad \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}\\
\Rightarrow \quad \cos \left( \alpha+\beta \right)&=\frac{33}{65}.\end{align} 

=====Question 4(iii)=====
If $\sin \alpha =-\dfrac{4}{5}$ and $\cos \beta =-\dfrac{12}{13}$,  $\alpha $in Quadrant III  and $\beta $in Quadrant II, find the exact value of  $\tan \left( \alpha +\beta  \right)$ .
====Solution====
Given: $\sin \alpha=-\dfrac{4}{5}$, $\alpha$ is in 3rd quadrant, \\
 $\sin \beta=-\dfrac{12}{13}$, $\beta$ is in 2nd quadrant.

We have an identity: $$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$ 
As $\alpha$ lies in 3rd quadrant and $\cos$ is -ive in 3rd quadrant,
\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 $$\sin \beta=\pm\sqrt{1-\cos^2\beta}.$$
As $\beta$ lies in 2nd quadrant and $\sin$ is +ive in 2nd quadrant,
\begin{align} &=\sqrt{1-{{\left(-\dfrac{12}{13} \right)}^{2}}}\\
&=\sqrt{1-\dfrac{144}{169}}=\sqrt{\dfrac{25}{169}}\\
\Rightarrow \quad \sin \beta &=\dfrac{5}{13}.\\
\tan \alpha &= \dfrac{\sin \alpha}{\cos \alpha}\\
&=\dfrac{\dfrac{-4}{5}}{\dfrac{-3}{5}}\\
\tan \alpha &=\dfrac{4}{3}\\
\tan \beta &= \dfrac{\sin \beta}{\cos \beta}\\
&=\dfrac{\dfrac{5}{13}}{\dfrac{-12}{13}}\\
\tan \beta &=\dfrac{-5}{12}.\end{align}

Now 
\begin{align}\tan \left( \alpha +\beta  \right)&=\dfrac{\tan \alpha+ \tan \beta}{1-\tan \alpha \tan \beta} \\ 
&=\dfrac{\dfrac{4}{3}+ \dfrac{-5}{12}}{1-(\dfrac{4}{3}) (\dfrac{-5}{12})}\\
&=\dfrac{\dfrac{4}{3}- \dfrac{5}{12}}{1+ \dfrac{20}{36}}=\dfrac{\dfrac{11}{12}}{ \dfrac{56}{36}}\\
\Rightarrow \quad \tan \left( \alpha+\beta \right)&=\dfrac{33}{56}.\end{align}

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