Question

secA-tanA=a+1/a-1 then cosA​

Answer 1

SecA-TanA=a+1/a-1

therefore,

1/cosA-secA/cosA=a+1/a-1

cosA-secA.CosA/CosA²=a+1/a-1

cosA²a-sec²A/Cos²A=a²+1/a²+1 by taking square on both side

1/cos²A=a²+1/a²+1

cos²A= 1

cosA=√1

cosA=1