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प्रश्न
Prove the following identity :
`(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ) = 2(1 + cotθ)`
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उत्तर
LHS = `(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ)`
= `((1 + sinθ)(cosecθ + cotθ) - (1 - sinθ)(cosecθ - cotθ))/(cosec^2θ - cot^2θ)`
= `(cosecθ + cotθ + 1 + cosθ - cosecθ + cotθ + 1 - cosθ)/(1 + cot^2θ - cot^2θ)` (∵ `cosec^2θ = 1 + cot^2θ`)
= 2 + 2cotθ = 2(1 + cotθ)
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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`
