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Hint: We will apply the formula of trigonometry, \[\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B\] to simplify the question and we will apply the formula \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] to finally find the value of \[\cos \left( \theta +\phi \right)\].

Complete step-by-step answer:

We are given the values of \[\sin \theta =\dfrac{3}{5}\] and \[\cos \phi =\dfrac{-12}{13}\]. Now we will apply the formula of \[\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B\] to the term \[\cos \left( \theta +\phi \right)\].

Therefore, we have \[\cos \left( \theta +\phi \right)=\cos \theta \cos \phi -\sin \theta \sin \phi \].

Clearly, we can notice that we know the values of \[\sin \theta =\dfrac{3}{5}\] and \[\cos \phi =\dfrac{-12}{13}\] and we are now to find the values of \[\sin \phi \] and \[\cos \theta \]. For this we will apply a trigonometric identity. Here we use the trigonometric identity given as, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. As we know the value of \[\sin \theta =\dfrac{3}{5}\] so by the identity formula we have,

\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1.....(1)\]

We will substitute the value of \[\sin \theta \] in equation (1). Thus we have \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\].

\[\Rightarrow {{\left( \dfrac{3}{5} \right)}^{2}}+{{\cos }^{2}}\theta =1\]

Now we will take the fraction \[{{\left( \dfrac{3}{5} \right)}^{2}}\]to the right side of the equal sign.

Therefore, we have \[{{\cos }^{2}}\theta =1-{{\left( \dfrac{3}{5} \right)}^{2}}\].

We know that \[{{\left( 3 \right)}^{2}}=9\] and \[{{\left( 5 \right)}^{2}}=25\]. Thus, we have that,

\[{{\cos }^{2}}\theta =1-\dfrac{9}{25}\]

Now we will take the LCM on \[=1-\dfrac{9}{25}\]. Since the LCM is clearly 25. Therefore, we have,

\[\begin{align}

& {{\cos }^{2}}\theta =\dfrac{25-9}{25} \\

& {{\cos }^{2}}\theta =\dfrac{16}{25} \\

& \cos \theta =\pm \sqrt{\dfrac{16}{25}} \\

\end{align}\]

As we know that the square root of 16 is 4 and the square root of 25 is 5. Thus, we get,

\[\cos \theta =\pm \dfrac{4}{5}\].

According to the question, we have that \[\theta \] and \[\phi \] lies in the second quadrant and we know that the value of \[\cos \theta \] in the second quadrant is negative. So, the value of \[\cos \theta =\dfrac{-4}{5}\] instead of \[\dfrac{4}{5}\].

Similarly, we know that the value of \[\cos \phi =\dfrac{-12}{13}\] and we need to find out the value of \[\sin \phi \] = ?

This can be done using the trigonometric identity \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. Therefore, we have \[{{\sin }^{2}}\phi +{{\cos }^{2}}\phi =1\].

Now we will substitute the value of \[\cos \phi =\dfrac{-12}{13}\] and we get,

\[si{{n}^{2}}\phi +{{\left( \dfrac{-12}{13} \right)}^{2}}=1\]

Now we take the value \[{{\left( \dfrac{-12}{13} \right)}^{2}}\] to the right side of the equality sign. Therefore, we have,

\[\begin{align}

& \Rightarrow si{{n}^{2}}\phi =1-{{\left( \dfrac{-12}{13} \right)}^{2}} \\

& \therefore si{{n}^{2}}\phi =1-\dfrac{144}{169} \\

\end{align}\]

This is because we know that the square of 12 = 144 and the square of 13 = 169. Now we take LCM here.

\[\begin{align}

& si{{n}^{2}}\phi =\dfrac{169-144}{169} \\

& si{{n}^{2}}\phi =\dfrac{25}{169} \\

& si{{n}^{2}}\phi =\pm \sqrt{\dfrac{25}{169}} \\

\end{align}\]

Since, we know that \[\sqrt{25}=5\] and \[\sqrt{169}=13\]. Thus we have,

\[si{{n}^{2}}\phi =\pm \dfrac{5}{13}\]

Since according to the question, we know that \[\phi \] lies in the second quadrant and in the second quadrant, sin is positive. Therefore, we have,

\[\sin \phi =\dfrac{5}{13}\]

Hence the value of equation,

\[\begin{align}

& \cos \left( \theta +\phi \right)=\cos \theta \cos \phi -\sin \theta \sin \phi \\

& \cos \left( \theta +\phi \right)=\left( \dfrac{-4}{5} \right)\left( \dfrac{-12}{13} \right)-\left(

\dfrac{3}{5} \right)\left( \dfrac{5}{13} \right) \\

& \cos \left( \theta +\phi \right)=\dfrac{48}{65}-\dfrac{15}{65} \\

& \cos \left( \theta +\phi \right)=\dfrac{33}{65} \\

\end{align}\]

Note: We can apply HCF and LCM method for finding out the square root of the numbers. For example for finding the square root of 169 we will factorize it as \[13\times 13\]. This will also be the case in LCM and since 13 is a number that multiplies by itself to give 169. So we can write,

\[\sqrt{169}=\sqrt{13\times 13}\] as \[\sqrt{169}\] = 13.

Complete step-by-step answer:

We are given the values of \[\sin \theta =\dfrac{3}{5}\] and \[\cos \phi =\dfrac{-12}{13}\]. Now we will apply the formula of \[\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B\] to the term \[\cos \left( \theta +\phi \right)\].

Therefore, we have \[\cos \left( \theta +\phi \right)=\cos \theta \cos \phi -\sin \theta \sin \phi \].

Clearly, we can notice that we know the values of \[\sin \theta =\dfrac{3}{5}\] and \[\cos \phi =\dfrac{-12}{13}\] and we are now to find the values of \[\sin \phi \] and \[\cos \theta \]. For this we will apply a trigonometric identity. Here we use the trigonometric identity given as, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. As we know the value of \[\sin \theta =\dfrac{3}{5}\] so by the identity formula we have,

\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1.....(1)\]

We will substitute the value of \[\sin \theta \] in equation (1). Thus we have \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\].

\[\Rightarrow {{\left( \dfrac{3}{5} \right)}^{2}}+{{\cos }^{2}}\theta =1\]

Now we will take the fraction \[{{\left( \dfrac{3}{5} \right)}^{2}}\]to the right side of the equal sign.

Therefore, we have \[{{\cos }^{2}}\theta =1-{{\left( \dfrac{3}{5} \right)}^{2}}\].

We know that \[{{\left( 3 \right)}^{2}}=9\] and \[{{\left( 5 \right)}^{2}}=25\]. Thus, we have that,

\[{{\cos }^{2}}\theta =1-\dfrac{9}{25}\]

Now we will take the LCM on \[=1-\dfrac{9}{25}\]. Since the LCM is clearly 25. Therefore, we have,

\[\begin{align}

& {{\cos }^{2}}\theta =\dfrac{25-9}{25} \\

& {{\cos }^{2}}\theta =\dfrac{16}{25} \\

& \cos \theta =\pm \sqrt{\dfrac{16}{25}} \\

\end{align}\]

As we know that the square root of 16 is 4 and the square root of 25 is 5. Thus, we get,

\[\cos \theta =\pm \dfrac{4}{5}\].

According to the question, we have that \[\theta \] and \[\phi \] lies in the second quadrant and we know that the value of \[\cos \theta \] in the second quadrant is negative. So, the value of \[\cos \theta =\dfrac{-4}{5}\] instead of \[\dfrac{4}{5}\].

Similarly, we know that the value of \[\cos \phi =\dfrac{-12}{13}\] and we need to find out the value of \[\sin \phi \] = ?

This can be done using the trigonometric identity \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. Therefore, we have \[{{\sin }^{2}}\phi +{{\cos }^{2}}\phi =1\].

Now we will substitute the value of \[\cos \phi =\dfrac{-12}{13}\] and we get,

\[si{{n}^{2}}\phi +{{\left( \dfrac{-12}{13} \right)}^{2}}=1\]

Now we take the value \[{{\left( \dfrac{-12}{13} \right)}^{2}}\] to the right side of the equality sign. Therefore, we have,

\[\begin{align}

& \Rightarrow si{{n}^{2}}\phi =1-{{\left( \dfrac{-12}{13} \right)}^{2}} \\

& \therefore si{{n}^{2}}\phi =1-\dfrac{144}{169} \\

\end{align}\]

This is because we know that the square of 12 = 144 and the square of 13 = 169. Now we take LCM here.

\[\begin{align}

& si{{n}^{2}}\phi =\dfrac{169-144}{169} \\

& si{{n}^{2}}\phi =\dfrac{25}{169} \\

& si{{n}^{2}}\phi =\pm \sqrt{\dfrac{25}{169}} \\

\end{align}\]

Since, we know that \[\sqrt{25}=5\] and \[\sqrt{169}=13\]. Thus we have,

\[si{{n}^{2}}\phi =\pm \dfrac{5}{13}\]

Since according to the question, we know that \[\phi \] lies in the second quadrant and in the second quadrant, sin is positive. Therefore, we have,

\[\sin \phi =\dfrac{5}{13}\]

Hence the value of equation,

\[\begin{align}

& \cos \left( \theta +\phi \right)=\cos \theta \cos \phi -\sin \theta \sin \phi \\

& \cos \left( \theta +\phi \right)=\left( \dfrac{-4}{5} \right)\left( \dfrac{-12}{13} \right)-\left(

\dfrac{3}{5} \right)\left( \dfrac{5}{13} \right) \\

& \cos \left( \theta +\phi \right)=\dfrac{48}{65}-\dfrac{15}{65} \\

& \cos \left( \theta +\phi \right)=\dfrac{33}{65} \\

\end{align}\]

Note: We can apply HCF and LCM method for finding out the square root of the numbers. For example for finding the square root of 169 we will factorize it as \[13\times 13\]. This will also be the case in LCM and since 13 is a number that multiplies by itself to give 169. So we can write,

\[\sqrt{169}=\sqrt{13\times 13}\] as \[\sqrt{169}\] = 13.

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