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प्रश्न
Prove the following trigonometric identities.
`(1 + cos theta - sin^2 theta)/(sin theta (1 + cos theta)) = cot theta`
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उत्तर
In the given question, we need to prove `(1 + cos theta - sin^2 theta)/(sin theta (1 + cos theta)) = cot theta`
Using the property `sin^2 theta + cot^2 theta = 1` we get
So
`(1 + cos theta - sin^2 theta)/(sin theta (1 + cos theta))`
`= (1 + cos theta - (1 - cos^2 theta))/(sin theta (1 + cos theta)`
`= (cos theta + cos^2 theta)/(sin theta (1 + cos theta))`
Solving further, we get
`(cos theta + cos^2 theta)/(sin(1 + cos theta)) = (cos theta (1 + cos theta))/(sin theta(1 + cos theta))`
`= cos theta/sin theta`
`= cot theta`
Hence proved.
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`square = 1 + tan^2θ` ...[Fundamental trigonometric identity]
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tan θ × `sqrt(1 - sin^2 θ)` is equal to:
Find the value of sin2θ + cos2θ
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`
