Question

prove that sin theta minus cos theta plus 1 ➗sin theta plus cos theta minus 1 is equal to 1 divided by sec theta minus tan theta

Answer 1

I hope this helps you

Answer 2

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

$Divide\: numerator \:and \:denominator \:by \\cos \theta ,we \:get$

$= \frac{\frac{(sin \theta -cos \theta + 1 )}{cos \theta</p><p>}}{\frac{sin \theta + cos \theta - 1 }{cos \theta}}$

$= \frac{ tan \theta - 1 + sec \theta }{ tan \theta + 1 - sec \theta }$

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

$\blue { ( By \: Trigonometric \:Identity )}$

$\boxed {\pink { sec^{2} \theta - tan^{2} \theta = 1 }}$

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

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

/* After cancellation , we get */

$= \frac{1}{sec\theta - tan \theta ) } \\= RHs$

$Hence \:proved.$

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