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प्रश्न
If cosθ = `5/13`, then find sinθ.
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उत्तर
cosθ = `5/13`
`sin^2θ + cos^2θ = 1`
`sin^2θ + (5/13)^2 = 1`
`sin^2θ = (1 - 25)/169`
`sin^2θ = (169 - 25)/169`
`sin^2θ = 144/169`
sinθ = `sqrt(144/169)`
sinθ = `12/13`
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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`
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Statement 2: cosec2θ + cot2θ = 1
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