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प्रश्न
Prove that:
`cot^2A/(cosecA - 1) - 1 = cosecA`
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उत्तर
`cot^2A/(cosecA - 1) - 1`
= `(cot^2A - cosecA + 1)/(cosecA - 1)`
= `(-cosecA + cosec^2A)/(cosecA - 1)`
= `(cosecA(cosecA - 1))/(cosecA - 1)`
= cosec A
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संबंधित प्रश्न
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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`
