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If 3 cot `theta = 2, `show that `((4 sin theta - 4 cos theta))/((2 sin theta + 6 cos theta ))=1/3`
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It is given that cos ЁЭЬГ = `2/3`
LHS = `( 4 sin theta -3 cos theta)/(2 sin theta + 6 cos theta)`
Dividing the above expression by sin ЁЭЬГ, ЁЭСдЁЭСТ ЁЭСФЁЭСТЁЭСб:
`(4-3 cot theta)/(2+6 cot theta) [тИ╡ cot theta = (cos theta)/(sin theta)]`
Now, substituting the values of cot ЁЭЬГ ЁЭСЦЁЭСЫ ЁЭСбтДОЁЭСТ ЁЭСОЁЭСПЁЭСЬЁЭСгЁЭСТ ЁЭСТЁЭСеЁЭСЭЁЭСЯЁЭСТЁЭСаЁЭСаЁЭСЦЁЭСЬЁЭСЫ, ЁЭСдЁЭСТ ЁЭСФЁЭСТЁЭСб:
`(4-3(2/3))/(2+6(2/3))`
=`(4-2)/(2+4)=2/6=1/3`
i.e., LHS = RHS
Hence proved.
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