Question

prove that tan theta +sec theta -1 upon tan theta -sec theta +1 = 1+sin theta upon cos theta

Answer 1

tan thita+sec thita-1/tan thita - sec thita + 1= 1 + sin thita / cos thita has been proved

Answer 2

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$\underline\bold{\huge{SOLUTION \: :}}$

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Here, i am holding the "THETAS" as "A".

So, we can proceed our solution as under :

$\boxed{L. H. S.}$

(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

= $\boxed{R. H. S.}$

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ULTIMATELY,

L.H.S. = R.H.S. (PROVED)

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