Question

If tanA =5/6 ,tan B =-1/3 ,then tanC( A+B) = ?
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Answer 1

$\huge\underline\mathrm\green{Question:-} \underline\green\Downarrow{ }$

$\pink\Longrightarrow{ }$If tanA =5/6 ,tan B =-1/3

ㅤㅤthen tan(A+B) = ?

$\Large\mathrm\red{Given\:☞}$

$\orange{\star{ }}$$\mathrm\orange{tan\:A\:=}$$\large\mathrm\orange{\frac{5}{6}}$

$\orange{\star{ }}$$\mathrm\orange{tan\:B\:=}$$\large\mathrm\orange{\frac{-1}{3}}$

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$\mathrm\blue{consider,tan\:(A\:+\:B)\:=}$$\Large\mathrm\blue{\frac{tan\:+\:tan\:B}{1\:-\:tan\:A\:tan\:B}}$

$\pink\Rightarrow{ }$$\mathrm\blue{tan\:(A\:+\:B)\:=}$$\Large\mathrm\blue{\frac{5/6\:+\:(-1/3)}{1\:-5/6 ×\:(-1/3)}}$

$\pink\Rightarrow{ }$$\mathrm\blue{tan\:(A\:+\:B)\:=}$$\Large\mathrm\blue{\frac{5/6\:-\:1/3}{1\:-(-5/18)}}$

$\pink\Rightarrow{ }$$\mathrm\blue{tan\:(A\:+\:B)\:=}$$\Large\mathrm\blue{\frac{1/2}{1\:+\:5/18}}$

$\pink\Rightarrow{ }$$\mathrm\blue{tan\:(A\:+\:B)\:=}$$\Large\mathrm\blue{\frac{1/2}{23/18}}$

$\pink\Rightarrow{ }$$\mathrm\blue{tan\:(A\:+\:B)\:=}$$\Large\mathrm\blue{\frac{1}{2}}$$\Large\mathrm\blue{×}$$\Large\mathrm\blue{\frac{18}{23}}$

$\pink\Rightarrow{ }$$\mathrm\blue{tan\:(A\:+\:B)\:=}$$\Large\mathrm\blue{\frac{9}{23}}$

$\rule{200pt}{3pt}$

$\mathrm\blue{Therefore,}$$\mathrm\blue{tan\:(A\:+\:B)\:=}$$\Large\mathrm\blue{\frac{9}{23}}$

Answer 2

Answer:

the answer is

$\frac{9}{23}$

Step-by-step explanation:

if

$\tan(A ) = \frac{5}{6} \\ \tan( B) = - \frac{1}{3}$

then we know the formula of tan(A+B) as

$tan(A+B) = \frac{tanA+ \tan B}{1 - tanA \tan B}$

Now putting on the values we get,

$tan(A+B) = \frac{tanA+ \tan B}{1 - tanA \tan B} \\ = \frac{ \frac{5}{6} + ( - \frac{1}{3}) }{1 - \frac{5}{6} ( - \frac{1}{3}) } \\ = \frac{ \frac{5}{6} - \frac{1}{3} }{1 + \frac{5}{18} } \\ = \frac{ \frac{5 - 2}{6} }{ \frac{18 + 5}{18} } \\ = \frac{ \frac{3}{6} }{ \frac{23}{18} } \\ = \frac{3}{6} \times \frac{18}{23} \\ = \frac{1}{2} \times \frac{18}{23} \\ = \frac{9}{23}$