Question

If 1+cos A =root 2 Then show that sec A -1=root 2

Answer 1

1+cosA=√2
or, cosA=√2-1
or, cosA=(√2-1)(√2+1)/(√2+1)
or, cosA=(2-1)/(√2+1)
or, cosA=1/(√2+1)
or, √2+1=1/cosaA
or, √2+1=secA
or, secA-1=√2 (proved)

Answer 2

1 + cosA = √2
CosA = √2 - 1

SecA = 1/CosA
          = $\frac{1}{\sqrt2 - 1}$
          = $\frac{1}{\sqrt2 - 1} . \frac{\sqrt2 + 1}{\sqrt2 + 1}$
          = $\sqrt2 + 1$
   SecA - 1 = $\sqrt2$