Question

Prove that 1 + secθ - tanθ / 1 + secθ + tanθ = 1 - sinθ / cosθ​

Answer 1

$\huge {\boxed {\mathrm {\purple{answer}}}}$

WE HAVE TO PROVE:-

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

so, L.H.S

$= > \frac{1 + \sin \theta - \tan \theta}{1 + \sin \theta + \tan \theta}$

$= > \frac{ {( \sec}^{2} \theta - { \tan }^{2} \theta) + \sec \theta - \tan \theta }{1 + \sec \theta + \tan \theta } \: \: \: \: [ \therefore { \sec }^{2} \theta - { \tan}^{2} \theta = 1 ]$

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

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

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

$= > \sec \theta - \tan \theta$

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

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

Hence,

$L.H.S=R.H.S$

Answer 2

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$\displaystyle \: \frac{1 + secθ - tanθ \: }{ 1 + secθ + tanθ}$

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

$= \displaystyle \: \frac{ (sec \theta - tan \theta + sec²\theta- tan²\theta}{ ( sec \theta +tan \theta + 1\: )}$

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

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

$\displaystyle \:= (sec \theta - tan \theta \: )$

$\displaystyle \: = \frac{( 1 - sin \theta ) }{cos \theta}$