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Question
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(cosec θ – cot θ)^2 = (1-cos theta)/(1 + cos theta)`
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Solution
L.H.S
= `(cosec θ – cot θ)^2`
= `(1/sintheta - costheta/sintheta)^2`
= `(1-costheta)^2/(sin^2 theta)`
= `(1-cos theta)^2/(1-cos^2theta)`
= `((1-costheta)(1-costheta))/((1-costheta)(1+cos theta)) `
= `(1-cos theta)/(1+costheta)`
= 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`
