Question

Prove that sinθ - cosθ + 1/(sinθ + cosθ -1) = 1/(secθ - tanθ)

Answer 1

Step-by-step explanation:

To Prove:-

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

Proof:-

$\rm LHS = \dfrac{sin\theta - cos\theta + 1}{sin\theta + cos \theta- 1}$

Dividing numerator and denominator with cosθ

$\rm = \dfrac{ \dfrac{sin \theta- cos \theta+ 1}{cos\theta} }{ \: \: \: \: \dfrac{ sin\theta + cos \theta- 1 }{cos\theta}\: \: \: \: }$

$\rm = \dfrac{ \dfrac{sin \theta}{cos\theta} - \dfrac{cos\theta}{cos\theta} + \dfrac{1}{cos\theta} }{ \: \: \: \: \dfrac{ sin\theta }{cos\theta} + \dfrac{cos\theta}{cos\theta} - \dfrac{1}{cos\theta} \: \: }$

$\rm = \dfrac{tan \theta - 1 + sec\theta}{tan \theta+ 1 - sec\theta}$

$\rm = \dfrac{tan \theta+ sec \theta- 1}{tan\theta - sec\theta + 1}$

Multiplying both the numerator and denominator with tanθ - secθ

$\rm = \dfrac{[ (tan \theta + sec\theta) - 1]tan \theta- sec \theta} {(tanθ - sec \theta+ 1)(tan\theta - sec\theta )}$

$\rm = \dfrac{ [(tan \theta + sec\theta) (tan\theta- sec \theta)] - 1(tan\theta- sec\theta )} {(tan\theta - sec\theta+ 1)(tan\theta - sec\theta )}$

$\rm = \dfrac{ tan ^{2} \theta - sec^{2}\theta - (tan \theta- sec \theta)} {(tan\theta - sec\theta + 1)(tan\theta - sec \theta)}$

$\rm = \dfrac{ - 1 - tan\theta + sec\theta} {(tan \theta- sec\theta + 1)(tan \theta- sec \theta)} \: \: \: \: \: \: \: \: \: \: \bigg( \because sec ^{2}\theta= 1 + tan\theta \dashrightarrow tan^{2}\theta - sec ^{2}\theta = - 1 \bigg)$

$\rm = \dfrac{ - (1 + tan\theta - sec\theta)} {(tan\theta- sec \theta+ 1)(tan \theta- sec\theta )}$

$\rm = \dfrac{ - \: \: \cancel{(1 + tan \theta - sec\theta)}} { \cancel{(tan \theta- sec \theta+ 1)} \: (tan\theta - sec \theta)}$

$\rm = \dfrac{ - 1} { \: tan\theta- sec\theta }$

$\rm = \dfrac{ 1} { \: sec\theta - tan \theta} = RHS$

LHS = RHS

Hence Proved