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प्रश्न
In tan θ = 1, find the value of 5cot2θ + sin2θ - 1.
उत्तर
Consider ΔABC, where ∠A = 90°
tan θ = `"Perpendicular"/"Basse" = "AB"/"AC" = 1 = (1)/(1)`
By Pythagoras theorem,
BC2
= AB2 + AC2
= 12 + 12
= 2
⇒ BC = `sqrt(2)`
Now,
cot θ = `(1)/"tan θ"` = 1
sin θ = `"AB"/"BC" = (1)/sqrt(2)`
∴ 5cot2θ + sin2θ - 1
= `5 xx (1)^2 + (1/sqrt(2))^2 - 1`
= `5 + (1)/(2) - 1`
= `4 + (1)/(2)`
= `(9)/(2)`.
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