Question

if cosec θ =13/12, find the value of 2sin θ-3cos θ/4sin θ-9cos θ

Answer 1

Aηѕωεɾ :-

Given :

• $\large\rm{Cosec \: θ = \frac{13}{12}}$

To Find :-

$\large \rm{ \dfrac{2 \: sin \: { \theta} \: - \: 3 \cos\: { \theta} }{4 \: sin \: { \theta} \: - \: 9 \: cos \: { \theta}} }$

Solution :-

$\bigstar \: \: \rm{ \cosec \: = \: \dfrac{13}{12} }$

${ \bigstar} \: \: \rm{sin \: = \dfrac{12}{13}}$

${ \bold{we \: knw \: that \:: \:}} \\ { \rm{ {sin \: { \theta}}^{2} + {cos \: { \theta}}^{2} }} = 1 \\ { \rm{{cos \: { \theta}}^{2}} = {1}^{2} - {sin \: { \theta}}^{2} } \\ { \rm{{cos \: { \theta}}^{2} = 1 - { (\dfrac{12}{13} )}^{2} }} \\ { \rm{{cos \: { \theta}}^{2} = 1 - \dfrac{144}{169} }} \\ { \rm{{cos \: { \theta}}^{2} = \: \frac{169 - 144}{169} }} \\ { \rm{{cos \: { \theta}}^{2} = \dfrac{25}{169} }} \\ { \rm{{cos \: { \theta}} = \sqrt{ \dfrac{25}{169} } }} \\{ \bigstar} \: \: { \rm{{cos \: { \theta}} = \dfrac{5}{13} }}$

$\rm{Now, \: Substitute \: </p><p>cosec \: = \dfrac{13}{12} \: ,</p><p>sin = \dfrac{12}{13} \: and \: </p><p>cos = \dfrac{5}{13} </p><p>\: into \: a \: Given \: Question -}$

$\implies \large\rm{ \dfrac{2 \: ( { \dfrac{12}{13} }) \: - \: 3 ( \dfrac{5}{13} ) }{4 \: ( \dfrac{12}{13} )\: - \: 9 \:( \dfrac{5}{13} )}}$

$\implies\large \rm{ \dfrac{\: ( { \frac{24}{13} }) \: - \: ( \dfrac{15}{13} ) }{ \: ( \dfrac{48}{13} )\: - \: \:( \dfrac{45}{13} )}}$

$\implies \large\rm{ \dfrac{\: { \dfrac{24 - 15}{13} }}{ \: \dfrac{48 - 45}{13} }}$

$\implies\large \rm{ \dfrac{\: { \dfrac{9}{13} }}{ \: \dfrac{3}{13} }}$

$\implies\large{ \rm{ \dfrac{9}{13} \times \dfrac{13}{3}}}$

$\implies\large{ \rm{ \dfrac{9}{3} }}$

$\implies\large{ \rm{ \blue{3}}}$

$\large{ \underline{ \rm{ \therefore} \:{the \: answer \: is \: 3}}} \: { \heartsuit}$