prove that tan theta +sec theta -1 upon tan theta -sec theta +1 = 1+sin theta upon cos theta
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tan thita+sec thita-1/tan thita - sec thita + 1= 1 + sin thita / cos thita has been proved
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yash510:
thnq so much
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Here, i am holding the "THETAS" as "A".
So, we can proceed our solution as under :
(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
=
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ULTIMATELY,
L.H.S. = R.H.S. (PROVED)
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Here, i am holding the "THETAS" as "A".
So, we can proceed our solution as under :
(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
=
==================================
ULTIMATELY,
L.H.S. = R.H.S. (PROVED)
==================================
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