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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

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3 years ago

``` = 1+sinA    cosA= 1+sinAx(1-sinA)    cosAx(1-sinA)=      1-sin2A          cosA(1-sinA)=       cos2A              cosA(1-sinA)=     cosA          1-sinAHence Proved
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3 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.
```
2 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
```
6 months ago
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