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Question
`(1+tan^2A)/(1+cot^2A)` = ______.
Options
sec2 A
−1
cot2 A
tan2 A
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Solution
`(1+tan^2A)/(1+cot^2A)` = tan2 A.
Explanation:
`(1+tan^2A)/(1+cot^2A) = (1+(sin^2A)/cos^2A)/(1+(cos^2A)/(sin^2A))`
= `((cos^2A + sin^2A)/cos^2A)/((sin^2A + cos^2A)/sin^2A)`
= `(1/cos^2A)/(1/sin^2A)`
= `(sin^2A)/cos^2A`
= `tan^2A`
Hence, alternative tan2 A is correct.
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Activity:
L.H.S. = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` ...`[sin^2"A" + square = 1]`
= `square` – cos2A ...[sin2A = 1 – cos2A]
= `square`
= R.H.S.
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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`
