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Question
Prove that `(sin theta)/(1-cottheta) + (cos theta)/(1 - tan theta) = cos theta + sin theta`
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Solution
L.H.S = `(sin theta)/(1 - cot theta) + (cos theta)/(1- tan theta)`
`= (sin theta)/(1 - cos theta/sin theta) + cos theta/(1 - sin theta/cos theta)`
`= sin^2 theta/(sin theta - cos theta) + cos^2 theta/(cos theta - sin theta)`
`= (sin^2 theta)/(sin theta - cos theta) - cos^2 theta/(sin theta - costheta)`
`= (sin^2 theta - cos^2 theta)/(sin theta - cos theta)`
`= ((sin theta - cos theta)(sin theta + cos theta))/(sin theta - cos theta)`
`= sin theta + cos theta`
= R.H.S
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
