Question

im confussed

Answer 1

Is there a certain part that you're confused with or is it the whole thing?

Answer 2

yes

Answer 3

I used to understand this stuff but not anymore. Your best bet is for one of actual smart people to get online.

@vocaloid @jhonyy9 @az etc

Answer 4

\[\tan \alpha=-\frac{ 12 }{ 5 }\]
\[\sec ^2\alpha-\tan ^2\alpha=1\]
\[\sec ^2\alpha=1+\tan ^2\alpha=1+\left( \frac{ 12 }{ 5 } \right)^2=1+\frac{ 144 }{ 25 }=\frac{ 25+144 }{ 25 }\]
\[=\frac{169}{25}\]
\[\sec \alpha=\pm \frac{ 13 }{ 5 }\]
\[\alpha~lies~\in~quadrant~2\]
\[\sec \alpha ~is~negative\]
\[\sec \alpha=-\frac{ 13 }{ 5 }\]
\[\cos \alpha=-\frac{ 5 }{ 13 }\]
\[\tan \alpha=\frac{ \sin \alpha }{ \cos \alpha}=-\frac{ 12 }{ 5 }\]
\[\sin \alpha=-\frac{ 12 }{ 5 }\times \frac{ -5 }{ 13 }=\frac{ 12 }{ 13 }\]
\[\cos \beta=\frac{ 3 }{ 5 }\]
\[\sin ^2\beta+\cos ^2\beta=1\]
\[\sin ^2\beta=1-\cos ^2\beta=1-\left( \frac{ 3 }{ 5} \right)^2=1-\frac{ 9 }{ 25 }\]
\[=\frac{ 25-9 }{ 25 }=\frac{ 16 }{ 25 }\]
\[\sin \beta=\pm \frac{ 4 }{ 5 }\]
\[\beta~lies~\in~quadrant~iv\]
\[\sin \beta ~is~negative.\]
\[\sin \beta=-\frac{ 4 }{5}\]
\[\cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta \]
put the values in the formula and complete it.

Answer 5

-63/65?

Answer 6

this is the hard way if i used a training;e and got those it wouldve saved a lot of time but thank you a lot !

Answer 7

\[\cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta \]
\[=-\frac{ 5 }{ 13 }\times \frac{ 3 }{ 5 }+\frac{ 12 }{ 13}\times \frac{ -4 }{ 5}\]
\[=-\frac{ 15 }{ 65 }-\frac{ 48 }{ 65 }\]
\[=-\frac{ 63 }{ 65 }\]
you are correct.