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Question
(sec θ + tan θ) . (sec θ – tan θ) = ?
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Solution
(sec θ + tan θ)(sec θ – tan θ)
= sec2θ – tan2θ ...[∵ (a + b)(a – b) = a2 – b2]
= 1 ...`[(∵ 1 + tan^2θ = sec^2θ),(∴ sec^2θ - tan^2θ = 1)]`
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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`
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