1.

Prove that : `(secA+1)/(tanA)=(tanA)/(secA-1)`

Answer» L.H.S. `=(secA+1)/(tanA)=(tanA+1)/(secA)xx(secA-1)/(secA-1)`
[divide numberator and denominator by (secA-1)]
`=(sec^(2)A-1)/(tanA(secA-1))=(tan^(2)A)/(tanA (secA-1))=(tanA)/(secA-1)=R.H.S`


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