Question

Prove that: sinA- cosA+1/ sinA + cosA -1 = 1/secA -tanA​

Answer 1

To prove :-

• $\sf \dfrac{sin \: A - cos \: A + 1}{sin \: A + cos \: A - 1} = \dfrac{1}{sec \: A - tan \: A}$

Prove :-

We have

• $\sf \dfrac{sin \: A - cos \: A + 1}{sin \: A + cos \: A - 1} = \dfrac{1}{sec \: A - tan \: A}$

Take L.H.S :

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

⠀⠀⠀⠀⠀

• First divide 1/cos A to both numerator and denominator.

⠀⠀⠀⠀⠀

$\implies \sf \dfrac{ \bigg( sin \: A - cos \: A + 1\bigg) \times \dfrac{1}{cos \: A}}{\bigg( sin \: A + cos \: A - 1 \bigg) \times \dfrac{1}{cos \: A}}$

⠀⠀⠀⠀⠀

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

⠀⠀⠀⠀⠀

• Sin A/cos A = tan A and 1/cos A = sec A

⠀⠀⠀⠀⠀

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

⠀⠀⠀⠀⠀

• Rearrange.

⠀⠀⠀⠀⠀⠀⠀⠀

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

⠀⠀⠀⠀⠀

• sec² A - tan² A = 1

⠀⠀⠀⠀⠀

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

⠀⠀⠀⠀⠀

• In denominator, take minus common from first bracket and use a² - b² = (a + b)(a - b) in second bracket.

⠀⠀⠀⠀

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

⠀

• Now, Take sec A - tan A common from denominator.

⠀⠀⠀⠀⠀

$\implies \sf \dfrac{\cancel{tan \: A + sec \: A - 1}}{(sec \: A - tan \: A) (\cancel{-1 + sec \: A + tan \: A})}$

⠀⠀⠀⠀⠀

$\implies \sf \dfrac{1}{sec \: A - tan \: A}$

⠀⠀⠀⠀⠀

$\implies \sf R.H.S$

Hence, Proved!!

Answer 2

${\large{\underline{\underline{\bf{Question:-}}}}}$

Q) Price that :

$\sf \dfrac{sin \: a - cos \: a + 1}{sin \: A + cos \: A- 1} = \dfrac{1}{sec \: A - tan \: A}$

${\large{\bf{\underline{\underline{Required\:Solution:-}}}}}$

Taking LHS

$\colon \rightarrow \sf \: \dfrac{sin \: A - cos \: A + 1}{sin \: A + cos \: A - 1}$

• Divide every term by “ cos A ” .

$\colon \rightarrow \sf \: \dfrac{ \dfrac{sin \: A}{cos \: A} - \dfrac{cos \: A}{cos \: A} + \dfrac{1}{cos \: A} }{ \dfrac{sin \: A}{cos \: A} + \dfrac{cos \: A}{cos \: A} - \dfrac{1}{cos \: A} }$

We know,

• $\rm \: \dfrac{sin \: \alpha }{cos \: \alpha } = tan \: \alpha$
• $\rm \: \dfrac{1}{cos \: \alpha } = sec \: \alpha$

Changing the terms which can be changed..

$\colon \rightarrow \sf \: \dfrac{tan \: A -1 + sec \: A}{tan \: A + 1 - sec \: A}$

Regrouping the terms ..

$\colon \rightarrow \sf \: \dfrac{tan \: A + sec \: A - 1}{tan \: A- sec \: A + 1}$

We know,

• $\rm \: 1 = {sec}^{2} \: \alpha - {tan}^{2} \: \alpha$

Change the “ 1 ” in denominator.

$\colon \rightarrow \sf \: \dfrac{tan \: A+ sec \: A - 1}{(tan \: A - sec \: A) + ( {sec}^{2} \: A - {tan}^{2} \: A)}$

Using

• x² - y² = (x+y) (x-y)

$\colon \rightarrow \sf \: \dfrac{tan \: A + sec \: A - 1}{(tan \: A - sec \:A) + (sec \: A + tan \: A)(sec \: A - tan \: A)}$

• Take minus (-) common from the first term of denominator.

$\colon \rightarrow \sf \: \dfrac{(tan \: A + sec \: A - 1)}{ - (sec \: A - tan \: A) + (sec \: A - tan \: A)(sec \: A + tan \: A)}$

• Take (sec A - tan A) common from the denominator.

$\colon \rightarrow \sf \: \dfrac{(tan \: A + sec \: A - 1)}{(sec \: A - tan \: A)( - 1 + sec \: A+ tan \: A)}$

• Rearranging the terms and cutting the unnecessary terms.

$\colon \rightarrow \sf \: \dfrac{ \cancel{(tan \: A + sec \: A - 1)}}{(sec \: A- tan \: A) \cancel{(tan \: A + sec \: A - 1)}}$

$\colon \rightarrow \red{ \sf \: \dfrac{1}{sec \: A - tan \: A} }$

• As you can see, it is equal to RHS

so,

$\star \: \boxed{ \pink{\tt{L.H.S.= R. H. S. }}} \: \therefore \green{ \bf \: Proved \: }$

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