Question

Help me pls! Thank you so much

Answer 1

`\bf cos\left[tan^(-1)\left((12)/(5) \right)+ tan^(-1)\left((-8)/(15) \right) \right]\\ \left. \qquad \qquad \quad \right.\uparrow \qquad \qquad \qquad \uparrow \\ \left. \qquad \qquad \quad \right.\alpha \qquad \qquad \qquad \beta \\\\\\ \textit{that simply means }tan(\alpha)=\cfrac{12}{5}\qquad and\qquad tan(\beta)=\cfrac{-8}{5} \\\\\\ \textit{so, we're really looking for }cos(\alpha+\beta)`

now.. hmmm -8/15 is rather ambiguous, since the negative sign is in front of the rational, and either 8 or 15 can be negative, now, we happen to choose the 8 to get the minus, but it could have been 8/-15

ok, well hmm so, the issue boils down to


`\bf tan(\theta)=\cfrac{opposite}{adjacent}\qquad thus \\\\\\ tan(\alpha)=\cfrac{12}{5}\cfrac{\leftarrow opposite=b}{\leftarrow adjacent=a} \\\\\\ \textit{so, what is the hypotenuse`


now, let's take a peek at the second angle, angle β


`\bf tan(\beta)=\cfrac{-8}{15}\cfrac{\leftarrow opposite=b}{\leftarrow adjacent=a} \\\\\\ \textit{again, let's find`

now, with that in mind, let's use the angle sum identity for cosine


`\bf cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}})\\\\ -----------------------------\\\\ cos({{ \alpha}} + {{ \beta}})= \left( \cfrac{5}{13} \right)\left( \cfrac{15}{17} \right)-\left( \cfrac{12}{13} \right)\left( \cfrac{-8}{17} \right) \\\\\\ cos({{ \alpha}} + {{ \beta}})= \cfrac{75}{221}-\cfrac{-96}{221}\implies cos({{ \alpha}} + {{ \beta}})= \cfrac{75}{221}+\cfrac{96}{221} \\\\\\ \boxed{cos({{ \alpha}} + {{ \beta}})=\cfrac{171}{221}}`