Question

if 4 cot A=3, find sin A +cos A/sin A-cos A,

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Answer 1

Answer:

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Answer 2

${\red{\huge{\underline{\mathbb{Given :-}}}}}$

$4cotA = 3 \\ \\ cotA = \frac{3}{4} = \frac{QR}{PQ} = \frac{base}{perpendicular} \\ \\ In \: \triangle \: PQR \implies \\ \\ By \: Pythagoras \: theorem :-\\ \\ ({PR})^{2} = ( {PQ})^{2} + ( {QR})^{2} \\ \\ ( {PR})^{2} = {4}^{2} + {5}^{2} \\ \\ ( {PR})^{2} = 16 + 9 \\ \\ ({PR})^{2} = 25 \\ \\ PR = \sqrt{25} \\ \\ PR = \sqrt{5 \times 5} \\ \\ PR = 5 \\ \\ {\purple{\boxed{\large{\bold{PR = 5}}}}} \\ \\ sin \theta= \frac{Perpendicular}{hypotenuse} \\ \\ sinA = \frac{PQ}{PR} = \frac{4}{5} \\ \\ cos \theta = \frac{Base}{Hypotenuse} \\ \\ cosA = \frac{QR}{PR} = \frac{3}{5} \\ \\ Putting ,\: \\ {\red{\boxed{\large{\bold{sinA = \frac{4}{5} \: and \: cosA = \frac{3}{5} }}}}} \\ \\ \frac{sinaA + cosA}{sinA - cosA} \\ \\ = \frac{ \frac{4}{5} + \frac{3}{5} }{ \frac{4}{5} - \frac{3}{5} } \\ \\ = \frac{ \frac{4 + 3}{5} }{ \frac{4 - 3}{5} } \\ \\ = \frac{4 + 3}{4 - 3} \\ \\ = 7 \\ \\ Therefore, </p><p>\\ {\green{\boxed{\large{\bold{ \frac{SinA + cosA}{sinA - cosA} = 7 }}}}}$

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