Question

Sin theta minus cos theta + 1 / sin theta + cos theta minus 1 is equal to 1 / secant theta minus tan theta

Answer 1

Question:

To Prove :

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

Step-by-step explanation:

L.H.S. = ${\sf{\ \ {\dfrac{sin \theta - cos \theta + 1}{sin \theta + cos \theta - 1}} }}$

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Dividing the numerator and denominator with cos θ, so as to get tan θ and sec θ.

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

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We can write this as :

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

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${\boxed{\tt{Identity \ : \ {\dfrac{sin \theta}{cos \theta}} = tan \theta}}}$

${\boxed{\tt{Identity \ : \ {\dfrac{1}{cos \theta}} = sec \theta}}}$

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

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Rearranging the terms in denominator.

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

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${\boxed{\tt{Identity \ : \ tan^2 \theta + 1 = sec^2 \theta}}}$

${\tt{From \ this, \ we \ get \ [ 1 = sec^2 \theta - tan^2 \theta ] }}$

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Putting this value of 1 in the denominator.

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

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We can write this as :

$\implies{\sf{ {\dfrac{tan \theta - 1 + sec \theta}{tan \theta - sec \theta + [ - (tan^2 \theta - sec^2 \theta)]}}}}$

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

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${\boxed{\tt{Identity \ : \ a^2 - b^2 = (a - b)(a + b)}}}$

${\tt{Here, \ a = tan \theta , \ b = sec \theta}}$

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

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Taking (tan θ - sec θ) common from the denominator.

$\implies{\sf{ {\dfrac{tan \theta - 1 + sec \theta }{ tan \theta - sec \theta [ 1 - (tan \theta + sec \theta)] }} }}$

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

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We can write this as :

$\implies{\sf{ {\dfrac{tan \theta - 1 + sec \theta }{ tan \theta - sec \theta [ - (tan \theta + sec \theta - 1) }} }}$

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Cancelling (tan θ + sec θ - 1) from both numerator and denominator.

$\implies{\sf{ {\dfrac{1}{(tan \theta - sec \theta)(- 1)}}}}$

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$\implies{\sf{ {\dfrac{1}{sec \theta - tan \theta}} }}$

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= R.H.S.

Hence, verified.