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प्रश्न
Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S. = `square`
= `cos^2θ xx square` ...`[1 + tan^2θ = square]`
= `(cos θ xx square)^2`
= 12
= 1
= R.H.S.
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उत्तर
L.H.S. = \[\boxed{\text{cos}^2θ · (1 + \text{tan}^2θ)}\]
= cos2θ × \[\boxed{\text{sec}^2θ}\] ...[1 + tan2θ = \[\boxed{\text{sec}^2θ}\]]
= (cos θ × \[\boxed{\text{sec} θ}\])2
= 12
= 1
= 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`
