Question

If tanA+secA=√2tanA.then show that secA—tanA=√secA

Answer 1

The above theory is wrong and so cannot be proved We know sec(^2)A-tan(^2)A=1 ------(1) Given tanA+secA=√2tanA=secA+tanA=√2tanA ---(2) We have to prove secA—tanA=√secA ----(3) sec(^2)A-tan(^2)A=(secA+tanA)(secA—tanA)=√2tanA*√secA =(√2sinA/√cosA)*(√1/√cosA) =√(2sinA)/cosA not equal to from (1) A trigonometric identity There above thing cannot be proved is wrong statement as it violates trigonometric identity