Question

15. If 3cot a = 2 then prove that
(4sina- 3cosa)/(2 sina + 6 cosa) =1/3

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Answer 1

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies 3 \: cot \: a = 2 \\ \\ \red{\underline \bold{To \: Prove :}} \\ \tt: \implies \frac{4 \: sin \: a - 3 \: cos \: a}{2 \: sin \: a + 6 \: cos \: a} = \frac{1}{3}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies 3 \: cot \: a = 2 \\ \\ \tt: \implies cot \: a = \frac{2}{3} - - - - - (1)\\ \\ \tt: \implies cot \: a = \frac{b}{p} - - - - - (2) \\ \\ \text{From \: (1) \: and \: (2)} \\ \tt: \implies p = 3 \\ \\ \tt: \implies b = 2 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {h}^{2} = {p}^{2} + {b}^{2} \\ \\ \tt: \implies {h}^{2} = {3}^{2} + {2}^{2} \\ \\ \tt: \implies {h}^{2} = 9 + 4 \\ \\ \tt: \implies h = \sqrt{13} \\ \\ \bold{For \: finding \: value } \\ \tt: \implies \frac{4 \: sin \: a - 3 \: cos \: a}{2 \: sin \: a + 6 \: cos \: a} = \frac{4 \times \frac{p}{h} - 3 \times \frac{b}{h} }{2 \times \frac{p}{h} + 6 \times \frac{b}{h} }$

$\tt: \implies \frac{4 \: sin \: a - 3 \: cos \: a}{2 \: sin \: a + 6 \: cos \: a} = \frac{4 \times \frac{3}{ \sqrt{13} } - 3 \times \frac{2}{ \sqrt{13} } }{2 \times \frac{3}{ \sqrt{13} } + 6 \times \frac{2}{ \sqrt{13} } } \\ \\ \tt: \implies \frac{4 \: sin \: a - 3 \: cos \: a}{2 \: sin \: a + 6 \: cos \: a} = \frac{ \frac{12 - 6}{ \sqrt{13} } }{ \frac{6 + 12}{ \sqrt{13} } } \\ \\ \tt: \implies \frac{4 \: sin \: a - 3 \: cos \: a}{2 \: sin \: a + 6 \: cos \: a} = \frac{6}{ \sqrt{13} } \times \frac{ \sqrt{13} }{18} \\ \\ \green{\tt: \implies \frac{4 \: sin \: a - 3 \: cos \: a}{2 \: sin \: a + 6 \: cos \: a} = \frac{1}{3} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \green{ \huge{ \boxed{ \tt proved}}}$