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प्रश्न
Prove the following:
`1 + (cot^2 alpha)/(1 + "cosec" alpha)` = cosec α
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उत्तर
L.H.S = `1 + (cot^2 alpha)/(1 + "cosec" alpha)`
= `1 + ((cos^2 alpha)/(sin^2 alpha))/((1 + 1)/(sin alpha))` ...`[∵ cot theta = (cos theta)/(sin theta) "and" "cosec" theta = 1/sin theta]`
= `1 + (cos^2 alpha)/(sinalpha (1 + sin alpha))`
= `(sin alpha(1 + sin alpha) + cos^2 alpha)/(sin alpha(1 + sin alpha))`
= `(sin alpha + (sin^2 alpha + cos^2 alpha))/(sin alpha(1 + sin alpha)` ...[∵ sin2θ + cos2θ = 1]
= `((sin alpha + 1))/(sin alpha(sin alpha + 1))`
= `1/sinalpha` ...`[∵ "cosec" theta = 1/sin theta]`
= cosec α
= 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`
