3cotA=2 find 2sinA-3cosA/2sinA+3cosA
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Given : cotA = 2/3
To Find:
2sinA-3cosA/2sinA+3cosA
divide numerator and denominator by sinA
{2(sinA/sinA) -3(cosA/sinA)}/{2(sinA/sinA) + 3(cosA/sinA)}
now (sinA/sinA) = 1
and (cosA/sinA) = cotA= 2/3
2-3cotA/2+3cotA
{2- 3*2/3}/{2+3*2/3}
2-2/2+2
=0
To Find:
2sinA-3cosA/2sinA+3cosA
divide numerator and denominator by sinA
{2(sinA/sinA) -3(cosA/sinA)}/{2(sinA/sinA) + 3(cosA/sinA)}
now (sinA/sinA) = 1
and (cosA/sinA) = cotA= 2/3
2-3cotA/2+3cotA
{2- 3*2/3}/{2+3*2/3}
2-2/2+2
=0
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