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Question
Prove the following trigonometric identities.
`(cos theta - sin theta + 1)/(cos theta + sin theta - 1) = cosec theta + cot theta`
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Solution
We have to prove the following identity
`(cos theta - sin theta + 1)/(cos theta + sin theta - 1) = cosec theta + cot theta`
Consider the LHS = `(cos theta - sin theta + 1)/(cos theta + sin theta - 1)`
`= (cos theta - sin theta + 1)/(cos theta + sin theta - 1) xx (cos theta + sin theta + 1)/(cos theta + sin theta + 1)`
`= ((cos theta + 1)^2 - (sin theta)^2)/((cos theta + sin theta)^2 - (1)^2)`
`= (cos^2 theta + 1 + 2 cos theta - sin^2 theta)/(cos^2 theta + sin^2 theta + 2 cos theta sin theta - 1)`
`= (cos^2 theta + 1 + 2 cos theta - (1 - cos^2 theta))/(1 + 2 cos theta sin theta - 1)`
`= (2 cos^2 theta + 2 cos theta)/(2 cos theta sin theta)`
`= (2 cos^2 theta + 2 cos theta)/(2 cos theta sin theta)`
`= (2 cos theta(cos theta + 1))/(2 cos theta sin theta)`
`= (cos theta + 1)/sin theta`
`= cos theta/sin theta + 1/sin theta`
`= cot theta + cosec theta`
= RHS
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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`
