Question

1+sec-tan/1+sec+tan=1-sin/cos

Answer 1

add theta ...as it is mandatory for applying trigonometric functions..

your answer is in pic attached..

hope helped.. ^_^

Answer 2

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TO PROVE

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

PROOF

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

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

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

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

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

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