prove that secA+tanA-1/tanA-secA+1 =cosA/1-sinA
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sec A + tanA - 1/tanA - secA +1 = cosA/1-sinA
Step-by-step explanation:
As we know,
sec^2 - tan^2= 1
= (secA+tanA-(sec^2 A - tan ^2A)) /tanA-secA+1)
We know also that,
a^2-b^2 = (a+b)(a-b)
By putting this,
= (sec A + tan A - (sec A + tan A) ( sec A - tan A) / ( tanA - secA+1)
= ( sec A + tan A ) ( 1- (sec A - tan A)/ ( tanA - secA+1)
= (sec A + tan A)/(+ tan A - sec A)/(tan A - sec A+ 1)
= (sec A + tan A)
We get,
= 1/cos A + sin A/cos A
= (1+ sin A) / cos A
= (1 + sin A )(1- sin A)/(cos A (1- sin A))
= (1- sin ^2 A/(cos A (1- sin A))
= cos ^2 A / (cos A (1- sin A))
= cos A /(1- sin A)
∵ proved.
Learn more: Sec A + tan A
brainly.in/question/19568554
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