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प्रश्न
\[\frac{\sin \theta}{1 + \cos \theta}\]is equal to
पर्याय
\[\frac{\sin \theta}{1 + \cos \theta}\]
\[\frac{1 - \cos \theta}{\cos \theta}\]
\[\frac{1 - \cos \theta}{\cos \theta}\]
\[\frac{1 - \sin \theta}{\cos \theta}\]
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उत्तर
The given expression is `sin θ/(1+cosθ)`
Multiplying both the numerator and denominator under the root by`(1-cosθ )` , we have
`sinθ/(1+cos θ)`
= `(sinθ (1-cos θ))/((1+cosθ)(1-cos θ))`
=`(sin θ(1-cos θ))/(1-cos^2 θ)`
= `(sin θ(1-cos θ))/sin^2 θ`
= `(1-cos θ)/sin θ`
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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`
