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Prove that tan a + sec a - 1/tan a - sec a + 1 =1 + sin a / cos a

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3 years ago

```							(tanA+secA-1)/(tanA-secA+1)=(1+sinA)/cos Amultiply LHS by cosA /cosA to get (sinA+1-cosA) / (sinA-1+cosA) multiply again by cosA/cosA to get (sinA.cosA+cosA-cos^2A) / cosA(sinA-1+cosA) = ( cosA(1+sinA) - (1-sin^2A) ) / cosA(sinA-1+cosA) = ( cosA(1+sinA) - (1+sinA)(1-sinA) ) / cosA(sinA-1+cosA) = ( (1+sinA)(cosA-1+sinA) ) / cosA(sinA-1+cosA) = (1+sinA)/cosA
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3 years ago
```							(tanA+secA-1)/(tanA-secA+1)=(1+sinA)/cos ARecall that: sec^2A-tan^2A = 1LHS = tanA+secA-(sec^2A-tan^2A)/tanA-secA+1​tanA+secA-[(secA+tanA)(secA-tanA)]/tanA-secA+1(tanA+secA)[1-(secA-tanA)]/tanA-secA+1 =tanA+secAtanA+secA = sinA/cosA+1/cosA = 1+sinA/cosA
```
2 years ago
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