Question

➟ Kindly explain your answer
➟ If you can't explain then don't answer this question!​

Answer 1

${\boxed{\underline{\tt{\orange{Required \: Answer:-}}}}}$

$\dag \: \: \: \sf{d.) \frac{ - 131}{109}}$

●GIVEN:-

• $\sf{tan \theta \: = \frac{ - 5}{12} }$
• $\sf{ \theta \: is \: not \: in \: fourth \: quadrant}$

●TO FIND:-

• $\sf{ \frac{tan(90 \degree \: + \theta) - sin(180 \degree - \theta)}{sin(270 \degree - \theta) + cosec(360 \degree - \theta} = }$

●FORMULA USED:-

• Pythagoras theorem
• $\sf{AB^2 +BC^2=AC^2}$

• $\sf{tan(90 \degree + \theta) = - cot \theta}$
• $\sf{sin(180 \degree - \theta) = sin \theta}$

• $\sf{sin(270 \degree - \theta) = - cos \theta}$
• $\sf{cosec(360 \degree - \theta) = - cosec\theta}$

●SOLUTION:-

$\sf{tan \theta= \frac{ - 5}{12} = \frac{perpendicular}{base} }$

According to Pythagoras theorem: -

$\sf{AB^2 +BC^2=AC^2} \\ \\ : \longrightarrow \sf{ {5}^{2} + {12}^{2} = {AC}^{2} } \\ \\ : \longrightarrow \sf{ {AC}^{2} = 25 + 144 } \\ \\ : \longrightarrow \sf{ AC = \sqrt{25 + 144} } \\ \\ : \longrightarrow \sf{ AC = \sqrt{169}} \\ \\ \rightarrow \sf \pink{AC =13}$

here, we are using the formulas:-

$\sf{ \frac{tan(90 \degree \: + \theta) - sin(180 \degree - \theta)}{sin(270 \degree - \theta) + cosec(360 \degree - \theta} = } \\ \\ \rightarrow \sf{ \frac{ - cot \theta - sin \theta}{ - cos \theta - cosec \theta} } \\ \\ \rightarrow \sf{ \frac{sin \theta \: + \: cot \theta}{cosec \theta + cos \theta} \: \: \: \: \: \: \: \: \: (1)}$

as tanis given in -ve, So it can be in II quadrant or IV quadrant.

but in question, theta is not in IV quadrant so it is in II quadrant.

In II quadrant, sin is +ve, and cos is -ve.

$: \longrightarrow \sf{sin \theta = \frac{5}{13} } \\ \\ : \longrightarrow \sf{cos \theta = \frac{ - 12}{13} } \\ \\ : \longrightarrow \sf{cosec \theta = \frac{13}{5} } \\ \\ : \longrightarrow \sf{cot \theta = \frac{ - 12}{5} }$

putting these values in the (1) equation

$: \longrightarrow \sf{ \frac{ \frac{5}{13} - \frac{12}{5} }{ \frac{ - 13}{5} - \frac{12}{ 13} } } \\ \\ \sf{lcm \: is \: 65} \\ \\ : \longrightarrow \sf{ \frac{ \frac{5 \times 5 - 12 \times 13}{ \cancel{65}} }{ \frac{ - 13 \times 5 - 12 \times 5}{ \cancel{65}} } } \\ \\ : \longrightarrow \sf{ \frac{25 - 156}{169 - 60} } \\ \\ : \longrightarrow \sf \huge \blue{ \frac{ - 131}{109} }$

So, The required answer is -131/109.