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If 3 `cot theta = 4 , "write the value of" ((2 cos theta - sin theta))/(( 4 cos theta - sin theta))`
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W e have ,
3 `cot theta = 4 `
⇒ ` cot theta = 4/3 `
Now,
`((2 cos theta + sin theta ))/((4 cos theta - sin theta))`
=` (((2 cos theta )/ sin theta + sin theta / sin theta))/(((4 cos theta) / sin theta - sin theta/ sin theta))` (ЁЭР╖ЁЭСЦЁЭСгЁЭСЦЁЭССЁЭСЦЁЭСЫЁЭСФ ЁЭСЫЁЭСвЁЭСЪЁЭСТЁЭСЯЁЭСОЁЭСбЁЭСЬЁЭСЯ ЁЭСОЁЭСЫЁЭСС ЁЭССЁЭСТЁЭСЫЁЭСЬЁЭСЪЁЭСЦЁЭСЫЁЭСОЁЭСбЁЭСЬЁЭСЯ ЁЭСПЁЭСж sin ЁЭЬГ)
=`((2 cot theta +1))/((4 cot theta -1))`
=`((2xx4/3 +1))/((4xx4/3-1))`
=`((8/3+1/1))/((16/3-1/1))`
=`(((8+3)/3))/(((16-3)/3))`
=`((11/3))/((13/3))`
=`11/13`
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