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प्रश्न
Prove the following trigonometric identities.
`cos theta/(1 + sin theta) = (1 - sin theta)/cos theta`
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उत्तर
We know that `sin^2 theta + cos^2 theta = 1`
Multiplying both numerator and the denominator by `(1 - sin theta)`, we have
`cos theta/(1 + sin theta) = (cos theta(1 - sin theta))/((1 + sin theta)(1 - sin theta))`
`= (cos theta(1 - sin theta))/(1 - sin^2 theta)`
`= (cos theta (1 - sin theta))/cos^2 theta`
`= (1 - sin theta)/cos theta`
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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`
