Question

Help me pls! Thank you so much

Answer 1

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

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 

$\bf tan(\theta)=\cfrac{opposite}{adjacent}\qquad thus \\\\\\ tan(\alpha)=\cfrac{12}{5}\cfrac{\leftarrow opposite=b}{\leftarrow adjacent=a} \\\\\\ \textit{so, what is the hypotenuse "c"?}\\ \textit{ well, let's use the pythagorean theorem} \\\\\\ c=\sqrt{a^2+b^2}\implies c=\sqrt{25+144}\implies c=\sqrt{169}\implies \boxed{c=13}\\\\ -----------------------------\\\\ \textit{this simply means }\boxed{cos(\alpha)=\cfrac{5}{13}\qquad \qquad sin(\alpha)=\cfrac{12}{13} }$


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

$\bf tan(\beta)=\cfrac{-8}{15}\cfrac{\leftarrow opposite=b}{\leftarrow adjacent=a} \\\\\\ \textit{again, let's find "c", or the hypotenuse} \\\\\\ c=\sqrt{15^2+(-8)^2}\implies c=\sqrt{289}\implies \boxed{c=17}\\\\ -----------------------------\\\\ thus\qquad \boxed{cos(\beta)=\cfrac{15}{17}\qquad \qquad sin(\beta)=\cfrac{-8}{17}}$

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

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