Question

if cot X -5/12 in second quadrant then find the value of trigonometric function

Answer 1

cot x=b/p=-5/12
h=13
tax =–12/5
sinx=p/h=13/5
cosecx =h/p=5/13
cos x=b/h=–12/13
secx =h/b=–13/5

Answer 2

$\red {\boxed{ \huge \star \: \mathbb{Hola! \: Mate \: \star}}}$
$\underline{\bf{QUESTION}} : \\ \\ \tiny\bf{if \: cot \: X \: \frac{ - 5}{12} \: in \: second \: quadrant \: then \: find \: the \: value \: of \: trigonometric} \\ \tiny \bf{ function.} \\ \\ \underline{\bf{SOLUTION}} : \\ \\ \bf{cot \: x = \frac{B}{P} \: = \frac{ - 5}{12} } \\ \\ \rightarrow\tiny \bf{by \: using \: pythagoras \: theorem > > } \\ \\ \rightarrow \bf{ {xy}^{2} = {yz}^{2} + {xz}^{2} } \\ \\ \bf{ {(xy)}^{2} = { - 5}^{2} + {12}^{2} } \\ \\ \bf{( {xy}^{2} ) = 25 + 144} \\ \\ \bf{ {(xy)}^{2} = 169 } \\ \\ \huge \underline{\bf{xy = 13}} \\ \\ \rightarrow\bf{other \: trigonometric \: ratios \: of \: x } \\ \rightarrow \bf{sin \: x = \frac{p}{h} = \frac{12}{13}} \\ \\ \rightarrow \bf{cos \: x = \frac{b}{h} = \frac{ - 5}{13} } \\ \\ \rightarrow \bf{tan \: x = \frac{p}{b} ={-} \frac { 12 } { 5} } \\ \\ \rightarrow \bf{sec \: x = \frac{h}{b} = {-}\frac{13}{ 5} } \\ \\ \rightarrow \bf{cosec \: x \: = \frac{h}{p} = \frac{13}{12} }$