Question

if cosec A=13/12 find the value of (2 sin A-3 cos A÷4 sin A -9 cos A)​

Answer 1

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Answer 2

SOLUTION

GIVEN

$\displaystyle \sf{ \cosec A = \frac{13}{12} }$

TO DETERMINE

$\displaystyle \sf{ \frac{2 \sin A - 3\cos A}{4 \sin A - 9\cos A} }$

EVALUATION

Here it is given that

$\displaystyle \sf{ \cosec A = \frac{13}{12} }$

We are aware of the Trigonometric identity that

$\displaystyle \sf{ { \cot}^{2} A = {\cosec }^{2} A - 1 }$

$\implies \displaystyle \sf{ { \cot}^{2} A = {\bigg( \frac{13}{12} \bigg) }^{2} - 1 }$

$\implies \displaystyle \sf{ { \cot}^{2} A = \frac{169}{144} - 1 }$

$\implies \displaystyle \sf{ { \cot}^{2} A = \frac{169 - 144}{144} }$

$\implies \displaystyle \sf{ { \cot}^{2} A = \frac{25}{144} }$

$\implies \displaystyle \sf{ { \cot} A = \frac{5}{12} }$

Now

$\displaystyle \sf{ \frac{2 \sin A - 3\cos A}{4 \sin A - 9\cos A} }$

Dividing Numerator and denominator both by SinA we get

$\displaystyle \sf{ \frac{2 \sin A - 3\cos A}{4 \sin A - 9\cos A} }$

$= \displaystyle \sf{ \frac{2 - 3\cot A}{4 - 9\cot A} }$

$= \displaystyle \sf{ \frac{2 - 3 \times \frac{5}{12} }{4 - 9 \times \frac{5}{12} } }$

$= \displaystyle \sf{ \frac{ \frac{24 - 15}{12} }{ \frac{48 - 45}{12} } }$

$= \displaystyle \sf{ \frac{ \frac{9}{12} }{ \frac{3}{12} } }$

$= \displaystyle \sf{ \frac{9}{12} \times \frac{12}{3} }$

$\sf{ = 3}$

FINAL ANSWER

$\boxed{ \: \: \displaystyle \sf{ \frac{2 \sin A - 3\cos A}{4 \sin A - 9\cos A} } = 3 \: \: }$

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