prove that tan theta upon sec theta -1 + tan theta upon sec theta + 1 = 2 cosec theta
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\underline\bold{\huge{SOLUTION \: :}}
SOLUTION:
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Here, i am holding the "THETAS" as "A".
So, we can proceed our solution as under :
\boxed{L. H. S.}
L.H.S.
(tan A + sec A - 1)/(tan A - sec A + 1)
= (tan A + sec A - sec²A + tan²A)/(tan A - sec A + 1)
= [tan A + sec A - {(sec A+tan A) (sec A - tan A)}]/[tan A - sec A + 1]
= [tan A + sec A (1 - sec A + tan A)]/(tan A - sec A + 1)
= tan A + sec A
= sin A/cos A + 1/cos A
= ( 1 + sin A ) / cos A
= \boxed{R. H. S.}
R.H.S.
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ULTIMATELY,
L.H.S. = R.H.S. (PROVED)
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