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प्रश्न
Prove the following identity :
sinθcotθ + sinθcosecθ = 1 + cosθ
उत्तर
sinθcotθ + sinθcosecθ = 1 + cosθ
LHS = `sinθcosθ/sinθ + sinθ1/sinθ`
= cosθ + 1 = RHS
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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`
