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Question
if `sin theta = 3/5 " evaluate " (cos theta - 1/(tan theta))/(2 cot theta)`
Solution
We have
`sin theta = 3/5`
In Δ ABC
`AC^2 = AB^2 + BC^2`
`=> (5)^2 = (3)^2 + (BC)^2`
`=> 25 = 9 + (BC)^2`
`=> (BC)^2 = 25 - 9`
=> (BC)^2 = 16
=> BC = 4
`:. cos theta = 4/5 and cot theta = 4/3`
Now `(cos theta - 1/(tan theta))/(2 cot theta) = ((4/5) - (4/3))/(2 xx 4/3)`
`= ((12 - 20)/15)/(8/3)`
`= (-8)/15 xx 3/8`
`= (-1)/5`
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