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Question
Prove that `(cosθ)/(1 + sinθ) = (1 - sinθ)/(cosθ)`.
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Solution
L.H.S. = `(cosθ)/(1 + sinθ)`
= `(cosθ)/(1 + sinθ) xx (1 - sinθ)/(1 - sinθ)` ...[On rationalising the denominator]
= `(cosθ(1 - sinθ))/(1 - sin^2θ)`
= `(cosθ(1 - sinθ))/(cos^2θ)` ...`[(∵ sin^2θ + cos^2θ = 1),(∴ 1 -sin^2θ = cos^2θ)]`
= `(1 - sinθ)/(cosθ)`
= R.H.S.
∴ `(cosθ)/(1 + sinθ) = (1 - sinθ)/(cosθ)`
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tan (90 – θ) = ?
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
