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प्रश्न
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
उत्तर
LHS = `((sinθ - cosθ + 1)/(sinθ + cosθ - 1)) xx ((sinθ + cosθ + 1)/(sinθ + cosθ + 1))`
LHS = `((sinθ + 1 - cosθ )/(sinθ + cosθ - 1)) xx ((sinθ + 1 + cosθ)/(sinθ + cosθ + 1))`
LHS = `((sinθ + 1)^2 - cos^2θ)/((sinθ + cosθ)^2 - 1^2)`
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Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
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