Question

$cosA=-\frac{12}{13}$ and $cotB=\frac{24}{7},$ 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)

Answer 1

We have,
$cosA=\frac{-12}{13}$ and $cotB=\frac{24}{7}$
$\therefore sinA=\sqrt{1-co{s}^{2}A}=\sqrt{1-{\left(\frac{-12}{13}\right)}^2}$
$=\sqrt{1-\frac{144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13}$
and,
$cosecB=-\sqrt{1+co{t}^{2}B}$
[$\because$ cosec is negative in third quadrant]
$=-\sqrt{1+{\left(\frac{24}{7}\right)}^2}=\sqrt{1+\frac{576}{49}}$
$=-\sqrt{\frac{49+576}{49}}=-\sqrt{\frac{625}{49}}=-\frac{25}{7}$
$\Rightarrow sinB=\frac{-7}{25}$ $[\because cosecB=\frac{1}{sinB}]$
Now,
$cosB=-\sqrt{1-si{n}^{2}B}$
[$\because$ $cos$ $\theta$ is negative in third quadrant]
$=-\sqrt{1-{\left(\frac{-7}{25}\right)}^2}=-\sqrt{1-\frac{49}{625}}$
$=-\sqrt{\frac{625-49}{625}}=-\sqrt{\frac{576}{625}}=\frac{-24}{25}$
Now,
$sin(A+B)=sinAcosB+cosAsinB$
$=\frac{5}{13}\times \left(\frac{-24}{25}\right)+\left(\frac{-12}{13}\right)\times \left(\frac{-7}{25}\right)$
$=\frac{-120}{325}+\frac{84}{325}$
$=\frac{-120+84}{325}$
$=\frac{-36}{325}$
(ii) We have,
$cosA=\frac{-12}{13}$ and $cotB=\frac{24}{7}$
$\therefore sinA=\sqrt{1-co{s}^{2}A}=\sqrt{1-{\left(\frac{-12}{13}\right)}^2}$
$=\sqrt{1-\frac{144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13}$
and,
$cosecB=-\sqrt{1+co{t}^{2}B}$
[$\because$ cosec is negative in third quadrant]
$=-\sqrt{1+{\left(\frac{24}{7}\right)}^2}=-\sqrt{1+\frac{576}{49}}$
$=-\sqrt{\frac{49+576}{49}}=-\sqrt{\frac{625}{49}}=-\frac{25}{7}$
$\Rightarrow sinB=\frac{-7}{25}$
Now,
$cosB=-\sqrt{1-si{n}^{2}B}$
[$\because$ cos is negative in third quadrant]
$-\sqrt{1-{\left(\frac{-7}{25}\right)}^2}=-\sqrt{1-\frac{49}{625}}$
$=-\sqrt{\frac{625-49}{625}}=-\sqrt{\frac{576}{625}}=\frac{-24}{25}$
Now,
$cos(A+B)=cosAcosB-sinAsinB$
$=\left(\frac{-12}{13}\right)\times \left(\frac{-24}{25}\right)-\left(\frac{5}{13}\right)\times \left(\frac{-7}{25}\right)$
$=\frac{288}{325}+\frac{35}{325}$
=$\frac{323}{325}$