====== Question 2, Exercise 10.2 ======
Solutions of Question 2 of Exercise 10.2 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 2(i)=====
If $\sin \theta =\dfrac{5}{13}$ and terminal ray of $\theta $ is in the second quadrant, then find $\sin 2\theta $.
====Solution==== 
Given: $\sin \theta =\dfrac{5}{13}$.

Using the identity
$$\cos \theta =\pm \sqrt{1-{{\sin }^{2}}\theta }.$$ 
As terminal ray of $\theta$ is in the second quadrant, value of $\cos$ is negative
\begin{align}\cos\theta &=-\sqrt{1-{{\sin }^{2}}\theta }\\
&=-\sqrt{1-\left(\frac{5}{13}\right)}\\
&=-\sqrt{\frac{144}{169}}\end{align}
$$\implies \cos\theta = -\dfrac{12}{13}$$
Thus, we have the following by using double angle identities:
\begin{align}\sin 2\theta &=2\sin \theta \cos \theta \\
&=2\left( -\dfrac{5}{13} \right)\left( \dfrac{12}{13} \right)\end{align}
$$\implies \bbox[4px,border:2px solid black]{\sin 2\theta=-\dfrac{120}{169}.}$$


=====Question 2(ii)=====
If $\sin \theta =\dfrac{5}{13}$ and terminal ray of $\theta $ is in the second quadrant, then find $\cos 2\theta $.
====Solution==== 
Given: $\sin \theta =\dfrac{5}{13}$.

Using the identity
$$\cos \theta =\pm \sqrt{1-{{\sin }^{2}}\theta }.$$ 
As terminal ray of $\theta$ is in the second quadrant, value of $\cos$ is negative
\begin{align}\cos\theta &=-\sqrt{1-{{\sin }^{2}}\theta }\\
&=-\sqrt{1-\left(\frac{5}{13}\right)}\\
&=-\sqrt{\frac{144}{169}}\end{align}
$$\implies \cos\theta = -\dfrac{12}{13}$$
Thus, we have the following by using double angle identities:
\begin{align}\cos 2\theta &={{\cos }^{2}}\theta -{{\sin }^{2}}\theta\\
&={{\left( \dfrac{12}{13} \right)}^{2}}-{{\left( \dfrac{-5}{13} \right)}^{2}}\\
&=\dfrac{144}{169}-\dfrac{25}{169}\end{align}
$$\implies \bbox[4px,border:2px solid black]{\cos 2\theta=-\dfrac{119}{169}.}$$


=====Question 2(iii)===== 
If $\sin \theta =\dfrac{5}{13}$ and terminal ray of $\theta $ is in the second quadrant, then find $\tan 2\theta $.
====Solution==== 
Given: $\sin \theta =\dfrac{5}{13}$.

Using the identity
$$\cos \theta =\pm \sqrt{1-{{\sin }^{2}}\theta }.$$ 
As terminal ray of $\theta$ is in the second quadrant, value of $\cos$ is negative
\begin{align}\cos\theta &=-\sqrt{1-{{\sin }^{2}}\theta }\\
&=-\sqrt{1-\left(\frac{5}{13}\right)}\\
&=-\sqrt{\frac{144}{169}}\end{align}
$$\implies \cos\theta = -\dfrac{12}{13}$$
Thus, we have the following by using double angle identities:
Thus, we have the following by using double angle identities.
\begin{align}\sin 2\theta &=2\sin \theta \cos \theta \\
&=2\left( -\dfrac{5}{13} \right)\left( \dfrac{12}{13} \right)=-\dfrac{120}{169}\end{align}
and
\begin{align}\cos 2\theta &={{\cos }^{2}}\theta -{{\sin }^{2}}\theta \\
&={{\left( \dfrac{12}{13} \right)}^{2}}-{{\left( \dfrac{-5}{13} \right)}^{2}}\\
&=\dfrac{144}{169}-\dfrac{25}{169}=\dfrac{119}{169}.\end{align}
Now 
\begin{align}\tan 2\theta &=\dfrac{\sin 2\theta }{\cos 2\theta }\\
&=\dfrac{\frac{-120}{169}}{\frac{119}{169}}\end{align}
$$\implies \bbox[4px,border:2px solid black]{\tan 2\theta=-\dfrac{120}{119}}$$

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