Question

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

Answer 1

$\huge\boxed{\underline{\underline{\green{Solution}}}}$

TO PROVE

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

PROOF

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

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

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

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

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

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

HENCE PROVED

Answer 2

Answer:

refer to the attachment

Step-by-step explanation:

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