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Question
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(tan theta)/(1-cot theta) + (cot theta)/(1-tan theta) = 1+secthetacosectheta`
[Hint: Write the expression in terms of sinθ and cosθ]
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Solution
L.H.S
= `(tantheta)/(1-cottheta) + (cottheta)/(1-tantheta) `
= `(sintheta/costheta)/(1-costheta/sintheta) + (costheta/sintheta)/(1-sintheta/costheta)`
= `(sintheta/costheta)/((sintheta-costheta)/(sintheta))+ (costheta/sintheta)/((costheta-sintheta)/costheta)`
= `(sin^2theta)/(costheta(sintheta-costheta)) - (cos^2theta)/(sintheta(sintheta-costheta))`
= `1/(sintheta - costheta)[(sin^2theta)/costheta - cos^2theta/sintheta]`
= `(1/(sintheta-costheta))[(sin^3theta-cos^3theta)/(sinthetacostheta)]`
= `(1/(sintheta-costheta))[((sintheta-costheta)(sin^2theta+cos^2theta+sinthetacostheta))/(sinthetacostheta)]`
= `((1+sinthetacostheta))/((sinthetacostheta))`
= sec θ cosec θ + 1
= 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`
