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Question
Write the value of cosec2 (90° − θ) − tan2 θ.
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Solution
We have,
`cosec^2 (90°-θ)- tan ^2θ= {cosec(90°-θ)}^2-tan ^2θ`
= `(secθ )^2-tan^2 θ`
= `sec^2 θ-tan ^2 θ`
We know that, ` sec^2 θ-tan ^2θ=1`
Therefore, \[{cosec}^2 \left( 90° - \theta \right) - \tan^2 \theta = 1\]
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Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
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Find the value of sin2θ + cos2θ
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`
