If cos A =

Solution:

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) $\tan (A+B)=\frac{\sin (A+B)}{\cos (A+B)}$

$=\frac{-36 / 325}{323 / 325}$

$=-\frac{36}{323}$