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Prove that (1+secA)/secA=sin^2A/(1-cosA)...

Prove that `(1+secA)/secA=sin^2A/(1-cosA)`.

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Given `1/(secA-tanA)-1/(cosA)=1/cosA-1/(secA-tanA)`
or `1/(secA-tanA)+1/(secA-tanA)=1/(cosA)+1/cosA`
Here `R.H.S.=2/cosA`
Now `L.H.S.=1/(secA-tanA)+1/(secA+tanA)`
`=(secA+tanA+secA-tanA)/((secA-tanA)(secA+tanA))`
`=2/cosA`
Thus, L.H.S.=R.H.S.
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CENGAGE PUBLICATION-TRIGONOMETRIC FUNCTIONS -SINGLE CORRECT ANSWER TYPE
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