Prove that , sin θ+cos θ−1sinθ−cos θ+1=sec θ−tanθ1

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Prove that , $\frac{sin θ - cos θ + 1}{sin θ + cos θ - 1} = \frac{1}{sec θ - tan θ}$

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Prove that ,

\frac{sin θ - cos θ + 1}{sin θ + cos θ - 1} = \frac{1}{sec θ - tan θ}

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sin θ - cos θ + 1 sin θ + cos θ - 1

= tan θ - 1 + sec θ tan θ + 1 - sec θ

= ( tan θ + sec θ ) - 1 (tan θ - sec θ) + 1

= {( tan θ + sec θ ) - 1 } ( tan θ - sec θ ) {(tan θ - sec θ) + 1} (tan θ - sec θ)

= ( tan ^{2} θ - sec ^{2} θ ) - ( tan θ - sec θ ) {tan θ - sec θ + 1} (tan θ - sec θ)

= - 1 tan θ - sec θ

= 1 sec θ - tan θ

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Question Text	Prove that , $\frac{sin θ - cos θ + 1}{sin θ + cos θ - 1} = \frac{1}{sec θ - tan θ}$
Answer Type	Text solution:1
Upvotes	150