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Question
Prove the following trigonometric identities.
sec A (1 − sin A) (sec A + tan A) = 1
Solution
We have to prove sec A(1 − sin A)(sec A + tan A) = 1
We know that sec2 A − tan2 A − 1
So,
sec A(1 − sin A)(sec A + tan A) = {sec A(1 − sin A)}(sec A + tan A)
= (sec A − sec A sin A)(sec A + tan A)
= `(sec A - 1/cos A sin A) (sec A + tan A)` ...`(∵ sec theta = 1/costheta)`
= `(sec A - sin A/cos A) (sec A + tan A)` ...`(∵ tan theta = sin theta/costheta)`
= (sec A − tan A)(sec A + tan A)
= sec2 A − tan2 A
= 1 = R.H.S. ... (∵ sec2 θ = 1 tan2 θ)
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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`
