If $\cos A=-\frac{12}{13}$ and $\cot B=\frac{24}{7}$, where $A$ lies in the second quadrant and $B$ in the third quadrant, find the values of the following:
(i) sin (A + B)
(ii) cos (A + B)
(iii) tan (A + B)
Given:
$\cos A=-\frac{12}{13}$ and $\cot B=\frac{24}{7}$
$A$ lies in the second quadrant and $B$ lies in the third quadrant.
We know that sine function is positive in the second quadrant and in the third quadrant, both $\sin \mathrm{e}$ and $\cos$ ine functions are negative.
Therefore,
$\sin A=\sqrt{1-\cos ^{2} A}=\sqrt{1-\left(\frac{-12}{13}\right)^{2}}=\sqrt{1-\frac{144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13}$
$\sin B=-\frac{1}{\sqrt{1+\cot ^{2} B}}=-\frac{1}{\sqrt{1+\left(\frac{24}{7}\right)^{2}}}=\frac{-1}{\sqrt{1+\frac{576}{49}}}=\frac{-1}{\sqrt{\frac{625}{49}}}=\frac{-7}{25}$
$\cos B=-\sqrt{1-\sin ^{2} B}=-\sqrt{1-\left(\frac{-7}{25}\right)^{2}}=-\sqrt{1-\frac{49}{625}}=-\sqrt{\frac{576}{625}}=-\frac{24}{25}$
Now,
(i) $\sin (A+B)=\sin A \cos B+\cos A+\sin B$
$=\frac{5}{13} \times \frac{-24}{25}+\frac{-12}{13} \times \frac{-7}{25}$
$=\frac{-120}{325}+\frac{84}{325}$
$=\frac{-36}{325}$
(ii) $\cos (A+B)=\cos A \cos B-\sin A \sin B$
$=\frac{-12}{13} \times \frac{-24}{25}-\frac{5}{13} \times \frac{-7}{25}$
$=\frac{288}{325}+\frac{35}{325}$
$=\frac{323}{325}$
(iii) $\tan (A+B)=\frac{\sin (A+B)}{\cos (A+B)}$
$=\frac{-36 / 325}{323 / 325}$
$=-\frac{36}{323}$
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