Question

prove that,
$\frac{ \sin \: theta \: - \cos \: theta + 1} { \sin \: theta \: + \cos \: theta - 1} = \frac{1}{ \sec \: theta - \tan \: theta }$
​

Answer 1

Step-by-step explanation:

$\bf{\underline{\underline\red{To\:Prove:-}}}$

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

$\bf{\underline{\underline\green{Proof:-}}}$

Let us prove by simplifying LHS and RHS separately.

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

By Rationalising the denominator:-

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

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

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

$\rm = \dfrac{ (sin \theta + 1) ^{2} -( cos \theta) ^{2} }{(sin \theta + cos \theta )^{2} - {(1)}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg( \because(a + b)(a - b) = {a}^{2} - {b}^{2} \bigg)$

$\rm = \dfrac{ sin^{2} \theta + 2sin \theta + 1 -cos ^{2} \theta }{sin^{2} \theta + cos ^{2} \theta + 2sin \theta cos \theta- 1}$

$\rm = \dfrac{ sin^{2} \theta + 2sin \theta + sin ^{2} \theta }{1 + 2sin \theta cos \theta- 1} \: \: \: \: \bigg( \because {sin}^{2} \theta + {cos}^{2} \theta = 1 \implies {sin}^{2} \theta = 1 - {cos}^{2} \theta \bigg)$

$\rm = \dfrac{ 2sin^{2} \theta + 2sin \theta }{ 2sin \theta cos \theta}$

$\rm = \dfrac{ 2sin \theta(sin \theta + 1) }{ 2sin \theta cos \theta}$

$\rm = \dfrac{ \cancel{ 2sin \theta} \: (sin \theta + 1) }{ \cancel{2sin \theta} \: cos \theta}$

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

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

$\rm = tan\theta + sec \theta$

$\rm = sec\theta + tan \theta$

By rationalising:-

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

$\rm = \dfrac{sec^{2} \theta - tan^{2} \theta}{sec \theta - tan \theta }$

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

LHS = RHS

Hence Proved