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

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

Grade:10

3 Answers

Yashaswi Agrawal
30 Points
6 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
5 years ago
Multiply ((tanA-1)+(secA))/ ((tanA+1)-secA)) by ((tanA-1)+(secA)) on numerator and denominator. Then use the identitype a2-b2 to proceed further. Thanks.
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

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



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