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