❏ Prove that:-
[tex] \sf{ \dfrac{ \sin \theta - \cos \theta + 1 }{ \sin \theta + \cos \theta - 1 } = \dfrac{1 + \sin \theta}{ \cos\theta } }[/tex]
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[tex] \sf{ \dfrac{ \sin \theta - \cos \theta + 1 }{ \sin \theta + \cos \theta - 1 } = \dfrac{1 + \sin \theta}{ \cos\theta } }[/tex]
Kindly don't spam please! And I really need help.. I've exam tomorrow T_T
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Verified answer
L H S :-
[tex]\: \: \: \: \: \: \: \: \: \rm{\dfrac{sin \theta - cos \theta + 1}{sin \theta + cos \theta - 1}}[/tex]
[tex]\: \: \: \: \: \: \: \: \: \rm{\dfrac{sin \theta + 1 - cos \theta}{sin \theta - 1 + cos \theta}}[/tex]
[tex]\: \: \: \: \: \: \: \rm{\dfrac{sin \theta + (1 - cos \theta)}{sin \theta - (1 - cos \theta)}}[/tex]
from rationalising the denominators
[tex]\: \: \: \: \: \rm{\dfrac{sin \theta + (1 - cos \theta)}{sin \theta - (1 - cos \theta)} \: \: \dfrac{sin \theta + (1 - cos \theta)}{sin \theta + (1 - cos \theta)}}[/tex]
[tex]\: \: \: \: \rm{\dfrac{[sin \theta + (1 - cos \theta)]^{2}}{[sin \theta - (1 - cos \theta)][sin \theta + (1 - cos \theta)]}}[/tex]
as we know -
[tex]\: \: \: \: \: \: \: \: \rm{\boxed{\bold{\blue{(a + b)^{2} \: \: = \: \: (a^{2} + 2ab + b^{2})}}}}[/tex]
[tex]\: \: \: \: \: \: \rm{\boxed{\bold{\blue{(a^{2} - b^{2}) \: \: = \: \: (a + b)(a - b)}}}}[/tex]
On applying these algebraic properties
[tex]\: \: \: \: \: \: \: \rm{\dfrac{sin^{2} \theta + (1 - cos \theta)^{2} + 2sin \theta(1 - cos \theta)}{(sin \theta)^{2} - (1 - cos \theta)^{2}}}[/tex]
[tex]\: \: \: \: \: \: \rm{\dfrac{sin^{2} \theta + 1 + cos^{2} \theta - 2cos \theta + 2sin \theta(1 - cos \theta)}{sin^{2} \theta - (1 + cos^{2} \theta - 2cos \theta)}}[/tex]
we also know that -
[tex]\: \: \: \: \: \: \: \: \: \: \rm{\bold{sin^{2} \theta + cos^{2} \theta \: \: = \: \: 1}}[/tex]
So -
[tex]\: \: \: \: \: \: \: \rm{\dfrac{1 + 1 - 2cos \theta + 2sin \theta(1 - cos \theta)}{sin^{2} \theta - 1 - cos^{2} \theta + 2cos \theta}}[/tex]
[tex]\: \: \: \: \: \: \: \: \: \rm{\dfrac{2 - 2cos \theta + 2sin \theta(1 - cos \theta)}{- 1 + sin^{2} \theta - cos^{2} \theta + 2cos \theta}}[/tex]
[tex]\: \: \: \: \: \: \: \rm{\dfrac{2(1 - cos \theta) + 2sin \theta(1 - cos \theta)}{- (1 - sin^{2} \theta) - cos^{2} \theta + 2cos \theta}}[/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \rm{\dfrac{2(1 - cos \theta)(1 + sin \theta)}{- cos^{2} \theta - cos^{2} \theta + 2cos \theta}}[/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \rm{\dfrac{2(1 - cos \theta)(1 + sin \theta)}{-2 cos^{2} \theta + 2cos \theta}}[/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \rm{\dfrac{\cancel{2}\cancel{(1 - cos \theta)}(1 + sin \theta)}{\cancel{2}cos \theta \cancel{(1 - cos \theta)}}}[/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm{\blue{\dfrac{(1 + sin \theta)}{cos \theta}}}[/tex]