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प्रश्न
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
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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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Solution:
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∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
