Question

tanΘ/1-tanΘ - cotΘ/1-cotΘ = cosΘ+sinΘ/cosΘ-sinΘ

Answer 1

Answer:

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Answer 2

Answer:

$\frac{tanθ}{1 - tanθ} - \frac{cotθ}{1 - cotθ} = \frac{ cosθ+sin θ}{cosθ -sin θ} \\ \frac{ \frac{sinθ}{cosθ} }{1 - \frac{sinθ}{cosθ} } - \frac{ \frac{cosθ}{sinθ} }{1 - \frac{cosθ}{sinθ} } = \frac{ cosθ+sin θ}{cosθ -sin θ} \\ \frac{ \frac{sinθ}{cosθ} }{ \frac{cosθ - sinθ}{cosθ} } - \frac{ \frac{cosθ}{sinθ} }{ \frac{sinθ - cosθ}{sinθ} } = \frac{ cosθ+sin θ}{cosθ -sin θ} \\ \frac{sinθ}{cosθ} \times \frac{cosθ}{cosθ - sinθ} - \frac{cosθ}{sinθ} \times \frac{sinθ}{sinθ - cosθ} = \frac{ cosθ+sin θ}{cosθ -sin θ} \\ \frac{sinθ}{cosθ -sin θ} - \frac{cosθ}{ sinθ-cosθ } = \frac{ cosθ+sin θ}{cosθ -sin θ} \\ \frac{sinθ}{cosθ - sinθ} + \frac{cosθ}{ cosθ-sin θ} =\frac{ cosθ+sin θ}{cosθ -sin θ} \\ \frac{sinθ +cosθ }{ cosθ- sinθ} =\frac{ cosθ+sin θ}{cosθ -sin θ} \\ \frac{ cosθ+sin θ}{cosθ -sinθ } = \frac{ cosθ+sin θ}{cosθ -sin θ} \\ LHS=RHS \\ Hence \: proved$