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प्रश्न
Prove the following identities:
`sinA/(1 - cosA) - cotA = cosecA`
उत्तर
`sinA/(1 - cosA) - cotA`
= `sinA/(1 - cosA) - cosA/sinA`
= `(sin^2A - cosA + cos^2A)/((1 - cosA)sinA)`
= `(1 - cosA)/((1 - cosA)sinA)`
= `1/sinA`
= cosec A
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संबंधित प्रश्न
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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`
