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# Prove that secA+ tanA-1÷ tanA- secA+1= cosA÷1- sinA

Yashaswi Agrawal
30 Points
3 years ago
= 1+sinA
cosA
= 1+sinAx(1-sinA)
cosAx(1-sinA)
=      1-sin2A
cosA(1-sinA)
cos2A
cosA(1-sinA)
=     cosA
1-sinA
Hence Proved
KaranKishoreDv
13 Points
3 years ago
Multiply ((tanA-1)+(secA))/ ((tanA+1)-secA)) by ((tanA-1)+(secA)) on numerator and denominator. Then use the identitype a2-bto proceed further. Thanks.
Rishi Sharma
one year ago
Dear Student,

secA+tanA-1)/tanA-secA+1)
We know that sec^2-tan^2=1
=(secA+tanA-(sec^2 A - tan ^2A)) /tanA-secA+1)
We also know that a^2-b^2=(a+b)(a-b)
= (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)
= 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)

Thanks and Regards