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प्रश्न
Show that `sqrt((1+cosA)/(1-cosA)) = cosec A + cot A`
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उत्तर
L.H.S = `sqrt((1-cosA)/(1+cos A))`
`= sqrt((1-cosA)/(1+cosA) xx (1 - cos A)/(1- cos A)) = sqrt((1- cosA)^2/(1-cos^2A))`
`=sqrt((1- cosA)^2/(sin^2A)) = (1-cosA)/sin A = 1/sin A - cos A/sin A = cosec A -cot A` = R.H.S
Hence prove.
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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`
