Question

Prove that: sin theta - cos theta + 1/ sin theta + cos theta - 1 = 1/ sec theta - tan theta using suitable identities




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Answer 1

$\bigstar \underline{\underline{\sf To\ prove:-}}$

• $\sf \frac{Sin\theta -cos\theta +1}{sin\theta +cos\theta -1}=\frac{1}{sec\theta - tan\theta}$

$\bigstar \underline{\underline{\sf Proof:-}}$

Formulas used:-

$\leadsto \sf \frac{sin\theta }{cos\theta } =tan\theta$

$\leadsto \sf \frac{1}{cos\theta } =sec\theta$

$\leadsto \sf sec^2\theta -tan^2\theta =1$

$\leadsto \sf a^2 -b^2 =(a+b)(a-b)$

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LHS:

$:\implies \sf \frac{ \sf sin\theta - cos\theta +1}{sin\theta +cos\theta -1}$

Dividing numerator & denominator by cosθ:

$:\implies \sf \frac{(sin\theta - cos\theta +1)/cos\theta}{(sin\theta +cos\theta -1)/cos\theta}$

$:\implies \sf \frac{(sin\theta /cos\theta-(cos\theta /cos\theta )+1/cos\theta}{(sin\theta /cos\theta ) +(cos\theta /cos\theta )-1/cos\theta }$

$:\implies \sf \frac{tan\theta -1 +sec\theta }{tan\theta +1-sec\theta }$

$:\implies \sf \frac{tan\theta +sec\theta -1}{tan\theta -sec\theta +(sec^2\theta - tan^2\theta )}$

$:\implies \sf \frac{tan\theta +sec\theta -1}{(tan\theta -sec\theta )+(sec\theta -tan\theta )(sec\theta +tan\theta )}$

$:\implies \sf \frac{\cancel{tan\theta +sec\theta -1}}{sec\theta -tan\theta \cancel{-1+sec\theta +tan\theta }}$

$:\implies \bold{\sf \frac{1}{sec\theta tan\theta }}$

Therefore,

$\mapsto \boxed{\boxed{\sf \frac{sin\theta - cos\theta + 1}{sin\theta + cos\theta - 1}= \frac{1}{sec\theta - tan\theta } }}$

Hence proved.

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Answer 2

$\bigstar \underline{\underline{\sf To\ prove:-}}$

$\sf \frac{Sin\theta -cos\theta +1}{sin\theta +cos\theta -1}=\frac{1}{sec\theta - tan\theta}$

$\bigstar \underline{\underline{\sf Proof:-}}$

Formulas used:-

$\leadsto \sf \frac{sin\theta }{cos\theta } =tan\theta$

$\leadsto \sf \frac{1}{cos\theta } =sec\theta$

$\leadsto \sf sec^2\theta -tan^2\theta =1$

$\leadsto \sf a^2 -b^2 =(a+b)(a-b)$

-----------------------

LHS:

$:\implies \sf \frac{ \sf sin\theta - cos\theta +1}{sin\theta +cos\theta -1}$

Dividing numerator & denominator by cosθ:

$:\implies \sf \frac{(sin\theta - cos\theta +1)/cos\theta}{(sin\theta +cos\theta -1)/cos\theta}$

$:\implies \sf \frac{(sin\theta /cos\theta-(cos\theta /cos\theta )+1/cos\theta}{(sin\theta /cos\theta ) +(cos\theta /cos\theta )-1/cos\theta }$

$:\implies \sf \frac{tan\theta -1 +sec\theta }{tan\theta +1-sec\theta }$

$:\implies \sf \frac{tan\theta +sec\theta -1}{tan\theta -sec\theta +(sec^2\theta - tan^2\theta )}$

$:\implies \sf \frac{tan\theta +sec\theta -1}{(tan\theta -sec\theta )+(sec\theta -tan\theta )(sec\theta +tan\theta )}$

$:\implies \sf \frac{\cancel{tan\theta +sec\theta -1}}{sec\theta -tan\theta \cancel{-1+sec\theta +tan\theta }}$

$:\implies \bold{\sf \frac{1}{sec\theta tan\theta }}$

Therefore,

$\mapsto \boxed{\boxed{\sf \frac{sin\theta - cos\theta + 1}{sin\theta + cos\theta - 1}= \frac{1}{sec\theta - tan\theta } }}$

Hence proved.

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HOPE IT HELPS!