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Question
Prove the following trigonometric identities.
`(1 + sin theta)/cos theta + cos theta/(1 + sin theta) = 2 sec theta`
Solution 1
We have to prove `(1 + sin theta)/cos theta + cos theta/1+ sin theta - 2 sec theta`
We know that, `sin^2 theta + cos^2 theta = 1`
Multiplying the denominator and numerator of the second term by `(1 - sin theta)` we have
`(1 + sin theta)/cos theta + cos theta/(1 + sin theta) = (1 = sin theta)/cos theta = (cos theta(1 - sin theta))/((1 + sin theta)(1 - sin theta))`
`= (1 + sin theta)/cos theta = (cos theta (1 - sin theta))/(1-sin theta)`
`= (1 + sin theta)/cos theta + (cos theta(1 - sin theta))/cos^2 theta`
`= (1 + sin theta)/cos theta + (1 - sin theta)/cos theta`
`= (1 + sin theta + 1 -sin theta)/cos theta`
`= 2/cos theta`
`= 2 sec theta`
Solution 2
LHS = `(1 + sin θ)/cos θ + cos θ/(1 + sin θ)`
= `(( 1 + sin θ)^2 + cos^2 θ)/(cos θ( 1 + sin θ))`
= `( 1 + sin^2 θ + 2 sin θ + cos^2 θ)/(cos θ( 1 + sin θ ))`
= `(1 + (sin^2θ + cos^2 θ) + 2 sin θ)/(cos θ(1 + sin θ))`
= `(1 + 1 + 2sin θ)/(cos θ(1 + sin θ))`
= `(2(1 + sin θ))/(cos θ(1 + sin θ))`
= 2 sec θ
Hence proved.
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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`
