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Question
Prove the following identities:
sec2A + cosec2A = sec2A . cosec2A
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Solution
L.H.S. = sec2A + cosec2A
= `1/(cos^2A) + 1/(sin^2A)`
= `(sin^2A + cos^2A)/(cos^2A sin^2A)`
= `1/(cos^2A sin^2A)`
= sec2A cosec2A
= R.H.S. ...(∵ sin2A + cos2A = 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`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
