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प्रश्न
Prove the following trigonometric identities.
`(1 - sin θ)/(1 + sin θ) = (sec θ - tan θ)^2`
Prove that:
`(1 - sin θ)/(1 + sin θ) = (sec θ - tan θ)^2`
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उत्तर
We have to prove `(1 - sin θ)/(1 + sin θ) = (sec θ - tan θ)^2`
We know that, sin2 θ + cos2 θ = 1
Multiplying both numerator and denominator by (1 − sin θ), we have
`(1 - sin θ)/(1 + sin θ) = ((1 - sin θ)(1 - sin θ))/((1 + sin θ)(1 - sin θ))`
`= (1 - sin θ)^2/(1 - sin^2 θ)`
`= ((1 - sin θ)/cos θ)^2`
`= (1/cos θ - sin θ/cos θ)^2`
`= (sec θ - tan θ)^2`
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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`
