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anonymous

  • 5 years ago

i need to find the tan(a-b) knowing tan a=5/12, sin $= -4/5 and sec B= 25/24 the restriction is (pi/2)<$<a<(3pi/2)<B<2pi. This is for precalc class and is kind of my last resort... I don't know how this site works actually, but please help ^^

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  1. anonymous
    • 5 years ago
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    $ is theta by the way.... ._.

  2. anonymous
    • 5 years ago
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    \[tan^{-1}(a-b)=\frac{tan(a)-tan(b)}{1+tan(a)tan(b)}\] so you need to find \[tan(a)\] and \[tan(b)\] you already have \[tan(a)=\frac{5}{12}\] \[sin(b)=-\frac{4}{5}\] \[cos(b)=\frac{24}{25}\] because secant is the reciprocal of cosine, so \[tan(b)=-\frac{5}{6}\] because it is sine over cosine. plug these numbers into the formula and you win.

  3. anonymous
    • 5 years ago
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    NO YOU WIN! XD

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